b: \(x^4+4\)

\(=x^4+4x^2+4-4x^2\)

\(=\left(x^2+2\right)^2-\left(2x\right)^2=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)

Ta có: \(\frac{x^2}{x^2+2x+2}+\frac{x^2}{x^2-2x+2}=\frac{5\left(x^2-5\right)}{x^4+4}+\frac{25}{4}\)

=>\(\frac{x^2\left(x^2-2x+2\right)+x^2\left(x^2+2x+2\right)}{\left(x^2+2x+2\right)\left(x^2-2x+2\right)}-\frac{5\left(x^2-5\right)}{\left(x^2+2x+2\right)\left(x^2-2x+2\right)}=\frac{25}{4}\)

=>\(\frac{x^4-2x^3+2x^2+x^4+2x^3+2x^2-5x^2+25}{x^4+4}=\frac{25}{4}\)

=>\(\frac{2x^4-x^2+25}{x^4+4}=\frac{25}{4}\)

=>\(25\left(x^4+4\right)=4\left(2x^4-x^2+25\right)\)

=>\(25x^4+100-8x^4+4x^2-100=0\)

=>\(17x^4+4x^2=0\)

=>\(x^2\left(17x^2+4\right)=0\)

=>\(x^2=0\)

=>x=0

a: \(x^4+4x^2+16\)

\(=x^4+8x^2+16-4x^2\)

\(=\left(x^2+4\right)_{}^2-\left(2x\right)^2=\left(x^2-2x+4\right)\cdot\left(x^2+2x+4\right)\)

\(\frac{x^2+2x}{\left(x+1\right)^2+3}-\frac{x^2-2x}{\left(x-1\right)^2+3}=\frac{16}{x^4+4x^2+16}\)

=>\(\frac{x^2+2x}{x^2+2x+4}-\frac{x^2-2x}{x^2-2x+4}=\frac{16}{\left(x^2+2x+4\right)\left(x^2-2x+4\right)}\)

=>\(\left(x^2+2x\right)\left(x^2-2x+4\right)-\left(x^2-2x\right)\left(x^2+2x+4_{}\right)=16\)

=>\(\left(x^2+2x\right)\left(x^2-2x\right)+4\left(x^2+2x\right)-\left(x^2-2x\right)\left(x^2+2x\right)-4\left(x^2-2x\right)=16\)

=>\(4\cdot\left(x^2+2x-x^2+2x\right)=16\)

=>4*4x=16

=>16x=16

=>x=1

b: =x-2

d: \(=-x^3+\dfrac{3}{2}-2x\)

Chịu

\(a,=\dfrac{x^2+4x+3-2x^2+2x+x^2-4x+3}{\left(x-3\right)\left(x+3\right)}=\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x-3}\\ b,=\dfrac{1-2x+3+2y+2x-4}{6x^3y}=\dfrac{2y}{6x^3y}=\dfrac{1}{x^2}\\ c,=\dfrac{75y^2+18xy+10x^2}{30x^2y^3}\\ d,=\dfrac{5x+8-x}{4x\left(x+2\right)}=\dfrac{4\left(x+2\right)}{4x\left(x+2\right)}=\dfrac{1}{x}\\ c,=\dfrac{x^2+2+2x-2-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{1}{x^2+x+1}\)

a: (2x+1)(3-x)(4-2x)=0

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

hay \(x\in\left\{-\dfrac{1}{2};3;2\right\}\)

b: 2x(x-3)+5(x-3)=0

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

=>x=3 hoặc x=-5/2

c: =>(x-2)(x+2)+(x-2)(2x-3)=0

=>(x-2)(x+2+2x-3)=0

=>(x-2)(3x-1)=0

=>x=2 hoặc x=1/3

d: =>(x-2)(x-3)=0

=>x=2 hoặc x=3

e: =>(2x+5+x+2)(2x+5-x-2)=0

=>(3x+7)(x+3)=0

=>x=-7/3 hoặc x=-3

f: \(\Leftrightarrow2x^3+5x^2-3x=0\)

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

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

hay \(x\in\left\{0;-3;\dfrac{1}{2}\right\}\)

Bạn chú ý đăng lẻ câu hỏi! 1/

a/ \(=x^3-2x^5\)

b/\(=5x^2+5-x^3-x\)

c/ \(=x^3+3x^2-4x-2x^2-6x+8=x^3=x^2-10x+8\)

d/ \(=x^2-x^3+4x-2x+2x^2-8=3x^2-x^3+2x-8\)

e/ \(=x^4-x^2+2x^3-2x\)

f/ \(=\left(6x^2+x-2\right)\left(3-x\right)=17x^2+5x-6-6x^3\)

cảm ơn bạn đã nhắc

a: \(\frac{x+1}{2x-6}-\frac{4}{2x-6}\)

\(=\frac{x+1-4}{2\left(x-3\right)}\)

\(=\frac{x-3}{2\left(x-3\right)}=\frac12\)

b: \(\frac{3x-4}{6x+3}-\frac{x-5}{6x+3}\)

\(=\frac{3x-4-x+5}{6x+3}\)

\(=\frac{2x+1}{3\left(2x+1\right)}=\frac13\)

c: \(\frac{x-1}{x-3}-\frac{3x-8}{3-x}+\frac{3-2x}{x-3}\)

\(=\frac{x-1}{x-3}+\frac{3x-8}{x-3}+\frac{3-2x}{x-3}\)

\(=\frac{x-1+3x-8+3-2x}{x-3}=\frac{2x-6}{x-3}=\frac{2\left(x-3\right)}{x-3}\)

=2

d: \(\frac{3}{x+5}-\frac{5}{x-7}\)

\(=\frac{3\left(x-7\right)-5\left(x+5\right)}{\left(x+5\right)\left(x-7\right)}=\frac{3x-21-5x-25}{\left(x+5\right)\left(x-7\right)}\)

\(=\frac{-2x-46}{\left(x+5\right)\left(x-7\right)}\)

e: \(\frac{3}{x+5}-\frac{5}{x-7}\)

\(=\frac{3\left(x-7\right)-5\left(x+5\right)}{\left(x+5\right)\left(x-7\right)}=\frac{3x-21-5x-25}{\left(x+5\right)\left(x-7\right)}\)

\(=\frac{-2x-46}{\left(x+5\right)\left(x-7\right)}\)

f: \(\frac{2}{x-2}+\frac{3}{x+2}+\frac{5x-18}{x^2-4}\)

\(=\frac{2}{x-2}+\frac{3}{x+2}+\frac{5x-18}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{2\left(x+2\right)+3\left(x-2\right)+5x-18}{\left(x-2\right)\left(x+2\right)}=\frac{2x+4+3x-6+5x-18}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{10x-20}{\left(x-2\right)\left(x+2\right)}=\frac{10}{x+2}\)

a, \(x^4-8x^2+16=\left(x^2-4\right)^2\)

b, \(\left(4x+5\right)^2-\left(5x+4\right)^2=\left(4x+5-5x-4\right)\left(4x+5+5x+4\right)=\left(1-x\right)\left(9x+9\right)=9\left(1-x\right)\left(1+x\right)=9\left(1-x^2\right)\)

c, \(\left(2x-3\right)^2-2\left(2x-3\right)\left(x+2\right)+\left(-x-2\right)^2=\left(2x-3-x-2\right)^2=\left(x-5\right)^2\)

a) \(x^4-8x^2+16=\left(x^2-4\right)^2\)

b) \(\left(4x+5\right)^2-\left(5x+4\right)^2=\left(4x+5-5x-4\right)\left(4x+5+5x+4\right)=9\left(1-x\right)\left(x+1\right)\)c) \(\left(2x-3\right)^2-2.\left(2x-3\right)\left(x+2\right)+\left(-x-2\right)^2=\left(2x-3-x-2\right)^2=\left(x-5\right)^2\)

Bài 1

A= (x-2)(2x-1)-2x(x+3)=2x²-x-4x+2-2x²-6x=-11x+2

Bài 1:

a) \(A=\left(x-2\right)\left(2x-1\right)-2x\left(x+3\right)\)

\(A=2x^2-x-4x+2-2x^2-6x\)

\(A=-11x+2\)

b) \(B=\left(3x-2\right)\left(2x+1\right)-\left(6x-1\right)\left(x+2\right)\)

\(B=6x^2+3x-4x-2-6x^2-12x+x+2\)

\(B=-12x\)

c) \(C=6x\left(2x+3\right)-\left(4x-1\right)\left(3x-2\right)\)

\(C=12x^2+18x-12x^2+8x+3x-2\)

\(C=29x-2\)

d) \(D=\left(2x+3\right)\left(5x-2\right)+\left(x+4\right)\left(2x-1\right)-6x\left(2x-3\right)\)

\(D=10x^2-4x+15x-6+2x^2-x+8x-4-12x^2+18x\)

\(D=36x-10\)

a) \(\left(3x-1\right)\left(2x+7\right)-\left(x+1\right)\left(6x-5\right)=16\)

\(\Leftrightarrow\left(6x^2+21x-2x-7\right)-\left(6x^2-5x+6x-5\right)-16=0\)

\(\Leftrightarrow6x^2+21x-2x-7-6x^2+5x-6x+5-16=0\)

\(\Leftrightarrow18x-18=0\)

\(\Leftrightarrow18x=18\)

\(\Leftrightarrow x=18:18\)

\(\Leftrightarrow x=1\)

Vậy \(x=1\)

b) \(\left(2x+3\right)^2-2\left(2x+3\right)\left(2x-5\right)+\left(2x-5\right)^2=x^2+6x+64\)

\(\Leftrightarrow\left[\left(2x+3\right)-\left(2x-5\right)\right]^2-\left(x^2+6x+64\right)=0\)

\(\Leftrightarrow\left(2x+3-2x+5\right)^2-x^2-6x-64=0\)

\(\Leftrightarrow8^2-x^2-6x-64=0\)

\(\Leftrightarrow64-x^2-6x-64=0\)

\(\Leftrightarrow-x^2-6x=0\)

\(\Leftrightarrow x\left(-x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\-x-6=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\-x=6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-6\end{matrix}\right.\)

Vậy \(x=0\) hoặc \(x=-6\)

a) \(\left(3x-1\right)\left(2x+7\right)-\left(x+1\right)\left(6x-5\right)=16\)

\(\Leftrightarrow\left(6x^2+21x-2x-7\right)-\left(6x^2-5x+6x-5\right)-16=0\)

\(\Leftrightarrow6x^2+21x-2x-7-6x^2+5x-6x+5-16=0\)

\(\Leftrightarrow18x-18=0\)

\(\Leftrightarrow18x=18\)

\(\Leftrightarrow x=18:18\)

\(\Leftrightarrow x=1\)

Vậy \(x=1\)

b, \(\left(2x+3\right)^2-2\left(2x+3\right)\left(2x-5\right)+\left(2x- 5\right)^2=x^2+6x+64\)

\(\Leftrightarrow\left[\left(2x+3\right)-\left(2x-5\right)\right]^2- \left(x^2+6x+64\right)=0\)

\(\Leftrightarrow\left(2x+3-2x+5\right)^2-x^2-6x-64=0\)

\(\Leftrightarrow8^2-x^2-6x-64=0\)

\(\Leftrightarrow64-x^2-6x-64=0\)

\(\Leftrightarrow-x^2-6x=0\)

\(\Leftrightarrow x\left(-x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\-x-6=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\-x=6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-6\end{matrix}\right.\)

Vậy \(x=0\) hoặc \(x=6\)