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đéo giúp hỏi ngu

a. \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{4}{x-2\sqrt{x}}\right)\cdot\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{4}{x-4}\right)\)

<=> \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{4}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\cdot\dfrac{\sqrt{x}-2+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

<=> \(P=\dfrac{x-4}{\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

<=> \(P=\dfrac{\sqrt{x}+2}{x-2\sqrt{x}}\)

b. Khi \(x=7+4\sqrt{3}=\left(2+\sqrt{3}\right)^2\) => \(\sqrt{x}=2+\sqrt{3}\)

=> \(P=\dfrac{2+\sqrt{3}+2}{7+4\sqrt{3}-2\left(2+\sqrt{3}\right)}=\dfrac{4+\sqrt{3}}{7+4\sqrt{3}-4-2\sqrt{3}}=\dfrac{4+\sqrt{3}}{3+2\sqrt{3}}=\dfrac{5\sqrt{3}-6}{3}\)

check giùm mik

a: Ta có: \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{4}{x-2\sqrt{x}}\right)\cdot\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{4}{x-4}\right)\)

\(=\dfrac{x-4}{\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}-2+4}{x-4}\)

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

b: Thay \(x=7+4\sqrt{3}\) vào P, ta được:

\(P=\dfrac{2+\sqrt{3}+2}{\sqrt{3}\left(2+\sqrt{3}\right)}=\dfrac{-6+5\sqrt{3}}{3}\)

a. \(\dfrac{1}{3}.\left(x-1\right)+\dfrac{2}{5}.\left(x+1\right)=0\)

=> \(\dfrac{1}{3}x-\dfrac{1}{3}+\dfrac{2}{5}x+\dfrac{2}{5}=0\)

=> \(\dfrac{1}{3}x+\dfrac{2}{5}x=0+\dfrac{1}{3}-\dfrac{2}{5}\)

=> \(\dfrac{11}{15}x=\dfrac{-1}{15}\)

=> \(x=\dfrac{-1}{11}\)

Đây toán 8 mà? :v

a,\(\dfrac{1}{5}x\left(x-1\right)+\dfrac{2}{5}x\left(x+1\right)=0\)

\(\Leftrightarrow5x\left(x-1\right)+6x\left(x+1\right)=0\)

\(\Leftrightarrow\left[5\left(x-1\right)+6x\left(x+1\right)\right]x=0\)

\(\Leftrightarrow\left(5x-5+6x+6\right)x=0\)

\(\Leftrightarrow\left(11+1\right)x=0\)

\(\Leftrightarrow11x+1=0;x=0\)

\(\Leftrightarrow x=-\dfrac{1}{11};x=0\)

Vậy....

bai nay de\

a) \(=x^3-\dfrac{1}{27}-x^2+\dfrac{2}{3}x-\dfrac{1}{9}=x^3-x^2+\dfrac{2}{3}x-\dfrac{2}{27}\)

b) \(=x^6-6x^4+12x^2-8-x^3+x+x^2-3x=x^6-6x^4-x^3+13x^2-2x-8\)

đấy nhá

b) Ta có: \(\left|x\right|-\dfrac{3}{4}=\dfrac{5}{3}\)

\(\Leftrightarrow\left|x\right|=\dfrac{5}{3}+\dfrac{3}{4}=\dfrac{20}{12}+\dfrac{9}{12}=\dfrac{29}{12}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{29}{12}\\x=-\dfrac{29}{12}\end{matrix}\right.\)

c) Ta có: \(\left|2x-\dfrac{1}{3}\right|+\dfrac{5}{6}=1\)

\(\Leftrightarrow\left|2x-\dfrac{1}{3}\right|=\dfrac{1}{6}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{1}{3}=\dfrac{1}{6}\\2x-\dfrac{1}{3}=\dfrac{-1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{1}{6}+\dfrac{1}{3}=\dfrac{1}{2}\\2x=\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=\dfrac{1}{12}\end{matrix}\right.\)

Ta có một số phân tích sau :  \(a^4\)\(+\)\(4\)\(=\)\(\left(a^2-2a+2\right)\)\(\left(a^2+2a+2\right)\)

Nhân mỗi biểu thức trong ngoặc ở cả tử thức với  \(16\)\(=\)\(2^4\), ta được :

\(A\)\(=\)\(\frac{\left(1+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right)\left(5^4+\frac{1}{4}\right)...\left(29^4+\frac{1}{4}\right)}{\left(2^4+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right)\left(6^4+\frac{1}{4}\right)...\left(30^4+\frac{1}{4}\right)}\)

\(A\)\(=\)\(\frac{\left(2^4+4\right)\left(6^4+4\right)\left(10^4+4\right)...\left(58^4+4\right)}{\left(4^4+4\right)\left(8^4+4\right)\left(12^4+4\right)...\left(60^4+4\right)}\)

Kết hợp với phân tích nêu trên, khi đó :

\(A\)\(=\)\(\frac{\left(2^2-2.2+2\right)\left(2^2+2.2+2\right)\left(6^2-2.6+2\right)\left(6^2+2.6+2\right)....\left(58^2-2.58+2\right)\left(58^2+2.58+2\right)}{\left(4^2-2.4+2\right)\left(4^2+2.4+2\right)\left(8^2-2.8+2\right)\left(8^2+2.8+2\right)....\left(60^2-2.60+2\right)\left(60^2+2.60+2\right)}\)

\(\Rightarrow\)\(A\)\(=\)\(\frac{2.10.26.50.82.122....3250.3482}{10.26.50.82.122....3482.3722}\)\(=\)\(\frac{2}{3722}\)\(=\)\(\frac{1}{1861}\)

đặt x+29=y

ta có :

\(\dfrac{1}{y^2}+\dfrac{1}{\left(y+1\right)^2}=\dfrac{5}{4}\)

\(\Leftrightarrow\dfrac{2y^2+2y+1}{y^4+2y^3+y^2}=\dfrac{5}{4}\)

\(\Leftrightarrow5y^4+10y^3-3y^2-8y-4=0\)

\(\Leftrightarrow\left(y-1\right)\left(5y^3+15y^2+12y+4\right)=0\)

\(\Leftrightarrow\left(y-1\right)\left(y+2\right)\left(5y^2+5y+2\right)=0\)

\(\left[{}\begin{matrix}y-1=0\Leftrightarrow y=1\Rightarrow x+29=1\Leftrightarrow x=-28\\y+2=0\Leftrightarrow y=-2\Rightarrow x+29=-2\Leftrightarrow x=-31\\5y^2+5y+2=0\left(loại\right)\end{matrix}\right.\)

vậy, S={-31;-28}