toán lớp 9 thì ai mà biết chỉ lớp 5 thôi

đáp án là : 0 bít !

sống bớt xàm đi bạn trẻ

\(\sqrt{3x^2-5x+1}-\sqrt{x^2-2}=\sqrt{3\left(x^2-x-1\right)}-\sqrt{x^2-3x+4}\)

\(4x^2-4-3x=\sqrt[3]{x^2\left(x^2-1\right)}\)

\(\Leftrightarrow4\left(x-1\right)\left(x+1\right)-3x=\sqrt[3]{x^2\left(x-1\right)\left(x+1\right)}\)

dat \(\left(x-1\right)\left(x+1\right)=y\)

\(4y-3x=\sqrt[3]{x^2y}\)

\(\Leftrightarrow\left(4y-3x\right)^3=x^2y\)

\(\Leftrightarrow64y^3-144y^2x+108yx^2-27x^3=x^2y\)

\(\Leftrightarrow64y^3-144y^2x+107yx^2-27x^3=0\)

\(\Leftrightarrow64y^3-64y^2x-80y^2x+80x^2y+27x^2y-27x^3=0\)

\(\Leftrightarrow\left(y-x\right)\left(64y^2-80xy+27x^2\right)=0\)

de thay \(64y^2-80xy+27x^2=\left(8y\right)^2-2.8y.5x+25x^2+2x^2=\left(8y-5x\right)^2+2x^2>0\)

\(\Rightarrow y=x\)hay \(\left(x-1\right)\left(x+1\right)=x\Rightarrow x^2-x-1=0\) 

\(\left(x-\frac{1}{2}\right)^2-\frac{5}{4}=0\Rightarrow\left(x-\frac{1}{2}\right)^2=\frac{5}{4}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{\sqrt{5}+1}{2}\\x=\frac{-\sqrt{5}+1}{2}\end{cases}}\)

câu b tương tự nhé bạn

\(\Leftrightarrow\left(3-x\right)\sqrt{x-1}+\sqrt{5-2x}=\sqrt{\left[\left(x-3\right)^2+1\right]\left(4-x\right)}\)

đặt 3-x=a;\(\sqrt{x-1}=b;\sqrt{5-2x}=c\Rightarrow b^2+c^2=4-x\)

\(\Leftrightarrow ab+c=\sqrt{\left(a^2+1\right)\left(b^2+c^2\right)}\)

<=>a²b²+2abc+c²=a²b²+b²+a²c²+c²

<=>b²-2abc+a²c²=0

<=>(b-ac)²=0

<=>b=ac

đến đây thì dễ rồi

\(\sqrt{x+8}=\sqrt{3x+2}+\sqrt{x+3}\) dkxd \(\left\{{}\begin{matrix}x\ge-8\\x\ge\\x\ge-\dfrac{2}{3}\end{matrix}\right.-3\)=>x\(\ge\)\(\dfrac{-2}{3}\)

\(x+8=3x+2+x+3+2\sqrt{\left(3x+2\right)\left(x+3\right)}\)

\(x+8=4x+5+2\sqrt{\left(3x+2\right)\left(x+3\right)}\)

\(x+8-4x-5=2\sqrt{\left(3x+2\right)\left(x+3\right)}\)

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

\(\left\{{}\begin{matrix}-3\left(x-3\right)\ge0\\\left(-3x+3\right)^2=4.\left(3x+2\right)\left(x+3\right)\end{matrix}\right.\)

Chắc tới đây bạn làm đc rồi nhỉ