Câu hỏi của Sakura

Question

1, gpt

a,$\sqrt{x^2-3x+2}+\sqrt{x+3}=\sqrt{x-2}+\sqrt{x^2+2x-3}$

b, $\left(4x+2\right)\sqrt{x+8}=3x^2+7x+8$

c,$\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=\sqrt{2}$

2/ cho x,y,z thỏa mãn : \(\...

Trần Thanh Phương · Accepted Answer

ráng làm nốt rồi đi ngủ thoyy

1.

a) ĐK: $x\ge2$

$\sqrt{x^2-3x+2}+\sqrt{x+3}=\sqrt{x-2}+\sqrt{x^2+2x-3}$

$\Leftrightarrow\sqrt{\left(x-1\right)\left(x-2\right)}+\sqrt{x+3}=\sqrt{x-2}+\sqrt{\left(x+3\right)\left(x-1\right)}$

$\Leftrightarrow\sqrt{\left(x-1\right)\left(x-2\right)}+\sqrt{x+3}-\sqrt{x-2}-\sqrt{\left(x+3\right)\left(x-1\right)}$

$\Leftrightarrow\sqrt{x-2}\left(\sqrt{x-1}-1\right)-\sqrt{x+3}\left(\sqrt{x-1}-1\right)=0$

$\Leftrightarrow\left(\sqrt{x-1}-1\right)\left(\sqrt{x-2}-\sqrt{x+3}\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=1\\sqrt{x-2}=\sqrt{x+3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-1=1\x-2=x+3\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x=2\x=\varnothing\end{matrix}\right.$

Vậy...

b) $\left(4x+2\right)\sqrt{x+8}=3x^2+7x+8$

$\Leftrightarrow2\left(2x+1\right)\sqrt{x+8}=4x^2+4x+1+x+8-x^2+2x-1$

$\Leftrightarrow2\left(2x+1\right)\sqrt{x+8}=\left(2x+1\right)^2+\left(x+8\right)-\left(x-1\right)^2$

$\Leftrightarrow\left(2x+1\right)^2-2\left(2x-1\right)\sqrt{x+8}+\left(x+8\right)-\left(x-1\right)^2=0$

$\Leftrightarrow\left(2x+1-\sqrt{x+8}\right)^2-\left(x-1\right)^2=0$

$\Leftrightarrow\left(2x+1-\sqrt{x+8}-x+1\right)\left(2x+1-\sqrt{x+8}+x-1\right)=0$

$\Leftrightarrow\left(x-\sqrt{x+8}+2\right)\left(3x-\sqrt{x+8}\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}x+2=\sqrt{x+8}\3x=\sqrt{x+8}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\x=1\end{matrix}\right.$$\Leftrightarrow x=1$

Vậy...

c) $\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=\sqrt{2}$

Nhân cả 2 vế với $\sqrt{2}$ ta được :

$pt\Leftrightarrow\sqrt{2x+2\sqrt{2x-1}}+\sqrt{2x-2\sqrt{2x-1}}=2$

$\Leftrightarrow\sqrt{\left(\sqrt{2x-1}+1\right)^2}+\sqrt{\left(\sqrt{2x-1}-1\right)^2}=2$

$\Leftrightarrow\left|\sqrt{2x-1}+1\right|+\left|\sqrt{2x-1}-1\right|=2$

Ta có : $\left|\sqrt{2x-1}+1\right|+\left|\sqrt{2x-1}-1\right|$

$=\left|\sqrt{2x-1}+1\right|+\left|1-\sqrt{2x-1}\right|\ge\left|\sqrt{2x-1}+1+1-\sqrt{2x-1}\right|=2$

Dấu "=" xảy ra $\Leftrightarrow\left(\sqrt{2x-1}+1\right)\left(1-\sqrt{2x-1}\right)\ge0\Leftrightarrow\frac{1}{2}\le x\le1$

2) $\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right):\frac{1}{x+y+z}=1$

$\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}$

$\Leftrightarrow\frac{1}{x}+\frac{1}{y}=\frac{1}{x+y+z}-\frac{1}{z}$

$\Leftrightarrow\frac{x+y}{xy}=\frac{z-x-y-z}{z\left(x+y+z\right)}$

$\Leftrightarrow\frac{x+y}{xy}=\frac{-\left(x+y\right)}{z\left(x+y+z\right)}$

$\Leftrightarrow z\left(x+y\right)\left(x+y+z\right)=-xy\cdot\left(x+y\right)$

$\Leftrightarrow\left(x+y\right)\left(xz+yz+z^2+xy\right)=0$

$\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}x+y=0\y+z=0\z+x=0\end{matrix}\right.$$\Leftrightarrow\left[{}\begin{matrix}x=-y\y=-z\z=-x\end{matrix}\right.$

TH1: $x=-y\Leftrightarrow x^{29}=-y^{29}\Leftrightarrow x^{29}+y^{29}=0$

Khi đó $B=0\cdot\left(x^{11}+y^{11}\right)\cdot\left(x^{2013}+y^{2013}\right)=0$

Tương tự 2 trường hợp còn lại ta đều được $B=0$

Vậy $B=0$Câu trả lời: ráng làm nốt rồi đi ngủ thoyy