\(a,VP=\dfrac{x\left(x+3\right)}{x\left(2x-5\right)}=\dfrac{x+3}{2x-5}=VT\\ b,VP=\dfrac{\left(3-x\right)^2}{\left(3-x\right)\left(3+x\right)}=\dfrac{3-x}{x+3}=VT\\ c,VP=\dfrac{\left(x+4\right)\left(x^2-4x+16\right)}{\left(3-x\right)\left(x^2-4x+16\right)}=\dfrac{x+4}{3-x}=VP\left(bạn.sửa.lại.đề.đi\right)\\ d,VT=\dfrac{x^3-2x^2+8x^2-16x+15x-30}{x^3-5x^2+8x^2-40x+15x-75}\\ =\dfrac{\left(x-2\right)\left(x^2+8x+15\right)}{\left(x-5\right)\left(x^2+8x+15\right)}=\dfrac{x-2}{x-5}=VP\)

\(f\left(x\right)+g\left(x\right)=6x^4-3x^2-5\\ f\left(x\right)-g\left(x\right)=4x^4-6x^3+7x^2+8x-9\\ \Rightarrow2f\left(x\right)=6x^4-3x^2-5+4x^4-6x^3+7x^2+8x-9\\ 2f\left(x\right)=10x^4-6x^3+4x^2+8x-14\\ 2f\left(x\right)=2\left(5x^4-3x^3+2x^2+4x-7\right)\\ \Rightarrow f\left(x\right)=5x^4-3x^3+2x^2+8x-14\)

\(f\left(x\right)+g\left(x\right)=6x^4-3x^2-5\\ \Rightarrow g\left(x\right)=6x^4-3x^2-5-f\left(x\right)\\ g\left(x\right)=6x^4-3x^2-5-5x^4+3x^3-2x^2-8x+14\\ g\left(x\right)=x^4+3x^3-5x^2-8x+9\)

câu d

\(D=\dfrac{\left(1-x^2\right)}{x}\left(\dfrac{x^2}{x+3}-1\right)+\dfrac{3x^2-14x+3}{x^2+3x}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{\left(1-x^2\right)\left(x^2-x-3\right)+3x^2-14x+3}{x\left(x+3\right)}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{x^2-x-3-x^4+x^3-3x^2+3x^2-14x+3}{x\left(x+3\right)}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{-x^4+x^3+x^2-15x}{x\left(x+3\right)}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{-x\left(x^3-x^2-x+15\right)}{x\left(x+3\right)}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{-\left(x^3-x^2-x+15\right)}{\left(x+3\right)}\end{matrix}\right.\)

Bài 1: 

a: Ta có: \(\sqrt{3x^2}=\sqrt{12}\)

\(\Leftrightarrow3x^2=12\)

\(\Leftrightarrow x^2=4\)

hay \(x\in\left\{2;-2\right\}\)

b: Ta có: \(\sqrt{\left(x-2\right)^2}=3\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=3\\x-2=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)

a) \(\left(x+2\right)\left(x^2-4x+4\right)-\left(x^3+2x^2\right)=5\)

\(\Leftrightarrow\left(x+2\right)\left(x^2-4x+4\right)-x^2\left(x+2\right)=5\)

\(\Leftrightarrow\left(x+2\right)\left(x^2-4x+4-x^2\right)=5\)

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

\(\Leftrightarrow4x-4x^2+8-8x=5\)

\(\Leftrightarrow-4x^2-4x+3=0\)

\(\Leftrightarrow4x^2+4x-3=0\)

\(\Leftrightarrow4x^2-2x+6x-3=0\)

\(\Leftrightarrow2x\left(2x-1\right)+3\left(2x-1\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\2x+3=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{3}{2}\end{matrix}\right.\)

Vậy \(x=\left\{\frac{1}{2};-\frac{3}{2}\right\}\)

b) \(6x^2-6x\left(-2+x\right)=36\)

\(\Leftrightarrow6x^2+12x-6x^2=36\)

\(\Leftrightarrow12x=36\)

\(\Leftrightarrow x=3\)

Vậy x = 3

c) \(\left(x+2\right)^2+\left(x-3\right)^2-2\left(x-1\right)\left(x+1\right)=9\)

\(\Leftrightarrow x^2+4x+4+x^2-6x+9-2\left(x^2-1\right)=9\)

\(\Leftrightarrow2x^2-2x+13-2x^2+2=9\)

\(\Leftrightarrow15-2x=9\)

\(\Leftrightarrow2x=6\)

\(\Leftrightarrow x=3\)

Vậy x = 3

d) \(\left(x+5\right)^2-9=0\)

\(\Leftrightarrow\left(x+5\right)^2=9\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x+5\right)^2=3^2\\\left(x+5\right)^2=\left(-3\right)^2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x+5=3\\x+5=-3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=-8\end{matrix}\right.\)

Vậy x ={-2; -8}

e) \(\left(x-2\right)^3=x^3+6x^2=7\) (Câu này sai đề thì phải! Mình sửa lại đề, có gì không giống với đề của bạn thì ib mình sửa nha!)

\(\left(x-2\right)^3-x^3+6x^2=7\)

\(\Leftrightarrow x^3-6x^2+12x-8-x^3+6x^2=7\)

\(\Leftrightarrow12x-8=7\)

\(\Leftrightarrow12x=15\)

\(\Leftrightarrow x=\frac{5}{4}\)

Vậy \(x=\frac{5}{4}\)

#Chúc bạn học tốt!

bạn giải rồi thi

bạn giải luôn rồi mà. Nhưng kết quả thì đúng đó

C

Đẳng thức nào đúng:

 A.(x-3)^2=x^2+6x+9                              B.(x-3)^2=9-x^2

C.(x-3)^2=9-6x+x^2                               D.(x-3)^2=x^2-3x+9

Câu C