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a: \(=\dfrac{4x^3+8x^2-11x+3-\left(x^2-5\right)\left(2x-1\right)-2x^3-5x^2+x+1}{\left(2x-1\right)^3}\)

\(=\dfrac{2x^3+3x^2-10x+4-2x^3+x^2+10x-5}{\left(2x-1\right)^3}\)

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

b: \(=\dfrac{1+x+1-x}{1-x^2}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(=\dfrac{2+2x^2+2-2x^2}{1-x^4}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(=\dfrac{4+4x^4+4-4x^4}{1-x^8}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(=\dfrac{8+8x^8+8-8x^8}{1-x^{16}}+\dfrac{16}{1+x^{16}}\)

\(=\dfrac{32}{1+x^{32}}\)

18 tháng 7 2018

1)3x(x-2)=7(x-2)

<=>3x(x-2)-7(x-2)=0

<=>(x-2)(3x-7)=0

x-2=0=>x=2

3x-7=0=>x=7/3

cn lại lm tg tự

18 tháng 7 2018

10)\(x^2-9x+20=0\)

\(\Leftrightarrow x^2-4x-5x+20=0\)

\(\Leftrightarrow x\left(x-4\right)-5\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x-5\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}x=4\\x=5\end{cases}}\)

a: \(\dfrac{96}{\left(x-4\right)\left(x+4\right)}+\dfrac{7+x}{4-x}=\dfrac{2x-1}{x+4}-3\)

\(\Leftrightarrow\dfrac{96}{\left(x-4\right)\left(x+4\right)}-\dfrac{\left(x+7\right)\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{\left(2x-1\right)\left(x-4\right)}{\left(x+4\right)\left(x-4\right)}-\dfrac{3\left(x-4\right)\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}\)

Suy ra: \(96-x^2-11x-28=2x^2-9x+4-3\left(x^2-16\right)\)

\(\Leftrightarrow-x^2-11x+68=2x^2-9x+4-3x^2+48\)

\(\Leftrightarrow-x^2-11x+68=-x^2-9x+52\)

=>-11x+68=-9x+52

=>-2x=-16

hay x=8(nhận)

b: \(\dfrac{2}{x-1}+\dfrac{3}{x-2}=\dfrac{3}{x-3}\)

\(\Leftrightarrow2\left(x-2\right)\left(x-3\right)+3\left(x-1\right)\left(x-3\right)=3\left(x-1\right)\left(x-2\right)\)

\(\Leftrightarrow2\left(x^2-5x+6\right)+3\left(x^2-4x+3\right)=3\left(x^2-3x+2\right)\)

\(\Leftrightarrow2x^2-10x+12+3x^2-12x+9=3x^2-9x+6\)

\(\Leftrightarrow5x^2-22x+21-3x^2+9x-6=0\)

\(\Leftrightarrow2x^2-13x+15=0\)

\(\Leftrightarrow2x^2-10x-3x+15=0\)

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

=>x=5(nhận) hoặc x=3/2(nhận)

10 tháng 5 2020

Bạn gõ ct lại dưới dạng latex (kí hiệu này gọi là latex Σ ) nhé

DD
13 tháng 8 2021

\(\left(1-x\right)\left(1+x+x^2+...+x^{31}\right)=1-x^{32}\)

\(\left(1-x\right)\left(1+x\right)\left(1+x^2\right)\left(1+x^4\right)\left(1+x^8\right)\left(1+x^{16}\right)\)

\(=\left(1-x^2\right)\left(1+x^2\right)\left(1+x^4\right)\left(1+x^8\right)\left(1+x^{16}\right)\)

\(=\left(1-x^4\right)\left(1+x^4\right)\left(1+x^8\right)\left(1+x^{16}\right)\)

\(=\left(1-x^8\right)\left(1+x^8\right)\left(1+x^{16}\right)\)

\(=\left(1-x^{16}\right)\left(1+x^{16}\right)\)

\(=1-x^{32}\)

Ta có đpcm.