đk: \(x\ne-2\)

\(PT\Leftrightarrow\frac{x^2}{\left(x+2\right)^2}=3x^2-6x-3\)

\(\Leftrightarrow x^2=\left(x^2+4x+4\right)\left(3x^2-6x-3\right)\)

\(\Leftrightarrow3x^4+6x^3-16x^2-36x-12=0\)

\(\Leftrightarrow\left(3x^4-18x^2\right)+\left(6x^3-36x\right)+\left(2x^2-12\right)=0\)

\(\Leftrightarrow3x^2\left(x^2-6\right)+6x\left(x^2-6\right)+2\left(x^2-6\right)=0\)

\(\Leftrightarrow\left(x^2-6\right)\left(3x^2+6x+2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2-6=0\\3x^2+6x+2=0\end{cases}}\Rightarrow x\in\left\{\pm\sqrt{6};\frac{-3\pm\sqrt{3}}{2}\right\}\)

bài yêu cầu gì vậy b?

(3x + 5)² - (2x + 1)² = 0

<=> (3x + 5 + 2x + 1)(3x + 5 - 2x - 1) = 0

<=> (5x + 6)(x + 4) = 0

<=> \(\orbr{\begin{cases}x=-\frac{6}{5}\\x=-4\end{cases}}\)

Vậy \(x\in\left\{-\frac{6}{5};-4\right\}\)là nghiệm phương trình

\(\left(3x+5\right)^2-\left(2x+1\right)^2=0\)

\(\Leftrightarrow\left(3x+5+2x+1\right)\left(3x+5-2x-1\right)=0\)

\(\Leftrightarrow\left(5x+6\right)\left(x+4\right)=0\Leftrightarrow x=-4;x=-\frac{6}{5}\)

Vậy tập nghiệm của phương trình là S = { -4 ; -6/5 } 

Trả lời :

= 123

T i c k nha !!

~HT~

Đáp án :

10 x 2 + 3 x 5 - 6 : 2 + 100 

= 132

Hok tốt

\(A=\sqrt{\frac{2+\sqrt{3}}{2}}-\sqrt{\frac{2-\sqrt{3}}{2}}\)

\(A=\sqrt{\frac{4+2\sqrt{3}}{4}}-\sqrt{\frac{4-2\sqrt{3}}{4}}\)

\(A=\sqrt{\frac{3+2\sqrt{3}+1}{4}}-\sqrt{\frac{3-2\sqrt{3}+1}{4}}\)

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

\(A=\left|\frac{\sqrt{3}+1}{2}\right|-\left|\frac{\sqrt{3}-1}{2}\right|\)

\(A=\frac{\sqrt{3}+1}{2}-\frac{\sqrt{3}-1}{2}\)

\(A=1\)

A=\(\sqrt{\frac{4+2\sqrt{3}}{4}}-\sqrt{\frac{4-2\sqrt{3}}{4}}=\)\(\sqrt{\frac{3+2\sqrt{3}+1}{4}}-\sqrt{\frac{3-2\sqrt{3+1}}{4}}\)=\(\frac{\sqrt{3}+1}{2}-\frac{\sqrt{3}-1}{2}=\frac{\sqrt{3}}{2}\)

\(\frac{x+2}{x+2}+\frac{5}{x-2}=\frac{4}{x^2-4}+1\)ĐK : \(x\ne\pm2\)

\(\Leftrightarrow\frac{\left(x+2\right)\left(x-2\right)+5\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}=\frac{4+\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)

\(\Rightarrow x^2-4+5x+10=4+x^2-4\Leftrightarrow5x+10=4\Leftrightarrow x=-\frac{6}{5}\)( tm )

ertgrrrr545454545454545454lo;ơ'n0u

3x - ( 2x - 5 ) = 10

3x - 2x + 5 = 10

x + 5 = 10

x = 10 - 5

x = 5

tôi nắm nay tôi 13 tuỏi chx gặp trường hợp là 3x-|2x-5|=10 toán lớp 8 nó phải là lớp 6

cho mình hình vẽ