minh lam cach nay ban xem thu xem sao:$x^{2}$-3x+k-1=0$\triangle$=9-4(k-1)=13-4kpt co nghiem$\Leftrightarrow$ $\triangle$$\geq$0 $\Leftrightarrow$k$\leq$$\frac{13}{4}$Ta co :$\left\{ \begin{array}{l} x_{1}+x_{2}=3 (1)\\ x_{1} .x_{2}=k-1 (2)\end{array} \right.$ket hop voi yeu cau bai ra , giai he$\Rightarrow$ka, theo bai ra ta co 2x1-5x2=-8 ket hop voi (1) ta duoc he:$\left\{ \begin{array}{l} x_{1}+x_{2}=3\\ 2x_{1}-5x_{2} =-8\end{array} \right.$$\Leftrightarrow$$\left\{ \begin{array}{l} x_{1}=1\\ x_{2}=2 \end{array} \right.$thay vao (2) ta duoc k=3b.theo bai ra ta co: $x^{2}_{1}$-$x^{2}_{2}$=-8 ket hop voi (1)ta co he:$\left\{ \begin{array}{l} (x_{1}-x_{2})(x_{1}+x_{2})=-8\\ x_{1}+x_{2}=3 \end{array} \right.$$\Leftrightarrow$$\left\{ \begin{array}{l} x_{1}=1/6\\ x_{2} =17/6\end{array} \right.$ thay vao (2)$\Rightarrow$k=53/36

minh lam cach nay ban xem thu xem sao:$x^{2}$-3x+k-1=0$\triangle$=9-4(k-1)=13-4kpt co nghiem$\Leftrightarrow$ $\triangle$$\geq$0 $\Leftrightarrow$k$\leq$$\frac{13}{4}$Ta co :$\left\{ \begin{array}{l} x_{1}+x_{2}=3 (1)\\ x_{1} .x_{2}=k-1 (2)\end{array} \right.$ket hop voi yeu cau bai ra , giai he$\Rightarrow$ka, theo bai ra ta co 2x1-5x2=-8 ket hop voi (1) ta duoc he:$\left\{ \begin{array}{l} x_{1}+x_{2}=3\\ 2x_{1}-5x_{2} =-8\end{array} \right.$$\Leftrightarrow$$\left\{ \begin{array}{l} x_{1}=1\\ x_{2}=2 \end{array} \right.$thay vao (2) ta duoc k=3b.theo bai ra ta co: $x^{2}_{1}$-$x^{2}_{2}$=-8 ket hop voi (1)ta co he:$\left\{ \begin{array}{l} (x_{1}-x_{2})(x_{1}+x_{2})=-8\\ x_{1}+x_{2}=3 \end{array} \right.$$\Leftrightarrow$$\left\{ \begin{array}{l} x_{1}=1/6\\ x_{2} =17/6\end{array} \right.$ thay vao (2)$\Rightarrow$k=53/36cau c lam tuong tu

minh lam cach nay ban xem thu xem sao:$x^{2}$-3x+k-1=0$\triangle$=9-4(k-1)=13-4kpt co nghiem$\Leftrightarrow$ $\triangle$$\geq$0 $\Leftrightarrow$k$\leq$$\frac{13}{4}$Ta co :$\left\{ \begin{array}{l} x_{1}+x_{2}=3 (1)\\ x_{1} .x_{2}=k-1 (2)\end{array} \right.$ket hop voi yeu cau bai ra , giai he$\Rightarrow$ka, theo bai ra ta co 2x1-5x2=-8 ket hop voi (1) ta duoc he:$\left\{ \begin{array}{l} x_{1}+x_{2}=3\\ 2x_{1}-5x_{2} =-8\end{array} \right.$$\Leftrightarrow$$\left\{ \begin{array}{l} x_{1}=1\\ x_{2}=2 \end{array} \right.$thay vao (2) ta duoc k=3

minh lam cach nay ban xem thu xem sao:$x^{2}$-3x+k-1=0$\triangle$=9-4(k-1)=13-4kpt co nghiem$\Leftrightarrow$ $\triangle$$\geq$0 $\Leftrightarrow$k$\leq$$\frac{13}{4}$Ta co :$\left\{ \begin{array}{l} x_{1}+x_{2}=3 (1)\\ x_{1} .x_{2}=k-1 (2)\end{array} \right.$ket hop voi yeu cau bai ra , giai he$\Rightarrow$ka, theo bai ra ta co 2x1-5x2=-8 ket hop voi (1) ta duoc he:$\left\{ \begin{array}{l} x_{1}+x_{2}=3\\ 2x_{1}-5x_{2} =-8\end{array} \right.$$\Leftrightarrow$$\left\{ \begin{array}{l} x_{1}=1\\ x_{2}=2 \end{array} \right.$thay vao (2) ta duoc k=3b.theo bai ra ta co: $x^{2}_{1}$-$x^{2}_{2}$=-8 ket hop voi (1)ta co he:$\left\{ \begin{array}{l} (x_{1}-x_{2})(x_{1}+x_{2})=-8\\ x_{1}+x_{2}=3 \end{array} \right.$$\Leftrightarrow$$\left\{ \begin{array}{l} x_{1}=1/6\\ x_{2} =17/6\end{array} \right.$ thay vao (2)$\Rightarrow$k=53/36

minh lam cach nay ban xem thu xem sao:$x^{2}$-3x+k-1=0$\triangle$=9-4(k-1)=13-4kpt co nghiem$\Leftrightarrow$ $\triangle$$\geq$0 $\Leftrightarrow$k$\leq$$\frac{13}{4}$Ta co :$\left\{ \begin{array}{l} x_{1}+x_{2}=3\\ x_{1} .x_{2}=k-1\end{array} \right.$ket hop voi yeu cau bai ra , giai he$\Rightarrow$k

minh lam cach nay ban xem thu xem sao:$x^{2}$-3x+k-1=0$\triangle$=9-4(k-1)=13-4kpt co nghiem$\Leftrightarrow$ $\triangle$$\geq$0 $\Leftrightarrow$k$\leq$$\frac{13}{4}$Ta co :$ \left\{ \begin{array}{l} x_{1}+x_{2}=3 (1)\\ x_{1} .x_{2}=k-1 (2)\end{array} \right.$ket hop voi yeu cau bai ra , giai he$\Rightarrow$ka, theo bai ra ta co 2x1-5x2=-8 ket hop voi (1) ta duoc he:$\left\{ \begin{array}{l} x_{1}+x_{2}=3\\ 2x_{1}-5x_{2} =-8\end{array} \right.$$\Leftrightarrow$$\left\{ \begin{array}{l} x_{1}=1\\ x_{2}=2 \end{array} \right.$thay vao (2) ta duoc k=3

minh lam cach nay ban xem thu xem sao:$x^{2}$-3x+k-1=0$\triangle$=9-4(k-1)=13-4kpt co nghiem$\Leftrightarrow$ $\triangle$$\geq$0 $\Leftrightarrow$k$\leq$$\frac{13}{4}$Ta co :$ \left\{ \begin{array}{l} x_{1}+x_{2}=\frac{3}{2}\\ x_{1} .x_{2}=k-1\end{array} \right.$ket hop voi yeu cau bai ra , giai he$\Rightarrow$k

minh lam cach nay ban xem thu xem sao:$x^{2}$-3x+k-1=0$\triangle$=9-4(k-1)=13-4kpt co nghiem$\Leftrightarrow$ $\triangle$$\geq$0 $\Leftrightarrow$k$\leq$$\frac{13}{4}$Ta co :$\left\{ \begin{array}{l} x_{1}+x_{2}=3\\ x_{1} .x_{2}=k-1\end{array} \right.$ket hop voi yeu cau bai ra , giai he$\Rightarrow$k

\left\{ $$\begin{array}{l}\sqrt{x-1}-x+3y-2\sqrt[3]{3y+2}=1\\ x+\sqrt{x-1}-3y-2\sqrt[3]{3y+2} =-7\end{array}$$ \right.\Leftrightarrow \left\{ $$\begin{array}{l} 2x-6y=-8\\ 2\sqrt{x-1}-4\sqrt[3]{3y+2}=-6\end{array}$$ \right.\Leftrightarrow \left\{ $$\begin{array}{l} x=3y-4 (1)\\ \sqrt{x-1} =\sqrt[3]{3y+2}-3 (2)\end{array}$$ \right. thay (1) vao` (2) ta co': \sqrt{3y-5}=\sqrt[3]{3y+2}-3Dat. t=\sqrt[3]{3y+2} giai tim` t\rightarrowy\rightarrowx

$\left\{ $$\begin{array}{l}\sqrt{x-1}-x+3y-2\sqrt[3]{3y+2}=1\\ x+\sqrt{x-1}-3y-2\sqrt[3]{3y+2} =-7\end{array}$$ \right$.\Leftrightarrow \left\{ $$\begin{array}{l} 2x-6y=-8\\ 2\sqrt{x-1}-4\sqrt[3]{3y+2}=-6\end{array}$$ \right.\Leftrightarrow \left\{ $$\begin{array}{l} x=3y-4 (1)\\ \sqrt{x-1} =\sqrt[3]{3y+2}-3 (2)\end{array}$$ \right. thay (1) vao` (2) ta co': \sqrt{3y-5}=\sqrt[3]{3y+2}-3Dat. t=\sqrt[3]{3y+2} giai tim` t\rightarrowy\rightarrowx