3(x+2)=-4(x-5)

3x+6=-4x+20

3x+4x=20-6

7x=14

x=14:7=2

ta có: 3/x - 5 = -4/x + 2

     => 3(x + 2) = -4(x + 5)

     => 3x + 6 = (-4)x + -20

     => 3x + 6 = (-4)x - 20

     => (-4)x - 3x = 20 + 6

     => (-1)x = 26

     => x = -26

khong biet

\(-4\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{6}\right)\le x\le-\frac{2}{3}.\left(\frac{1}{3}-\frac{1}{2}-\frac{3}{4}\right)\)

\(\Rightarrow-\frac{13}{3}.\left(\frac{3}{6}-\frac{1}{6}\right)\le x\le-\frac{2}{3}.\left(\frac{4}{12}-\frac{6}{12}-\frac{9}{12}\right)\)

\(\Rightarrow-\frac{13}{3}.\frac{2}{6}\le x\le-\frac{2}{3}.\frac{-11}{12}\)

\(\Rightarrow\frac{-13}{9}\le x\le\frac{11}{18}\)

\(\Rightarrow\frac{-26}{18}\le x\le\frac{11}{18}\)

=> -1,44444444444........... ≤ x ≤ 0,6111111111...........

Mà x ∈ Z

=> x ∈ { -1 ; 0 }

\(x\in\varnothing\) 

\(\frac{x+2}{x-2}-\frac{x-2}{x+2}+\frac{16}{4-x^2}\)

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

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

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

\(=\frac{8x-16}{\left(x-2\right)\left(x+2\right)}\)

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

\(=\frac{8}{x+2}\)

bạn ơi, cho mình hỏi

4-x² thì sẽ = 2²-x² = (2-x)(2+x)

vậy sao bạn ghi là (x-2)(x+2) vậy???

Biến đổi như thế này nha bạn :

\(\frac{16}{4-x^2}=\frac{-1\cdot16}{-1\cdot\left(4-x^2\right)}\)

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

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

Bạn tự tìm ĐKXĐ nhé :)

Xét tử thức : \(\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}=\sqrt{x-4+4\sqrt{x-4}+4}+\sqrt{x-4-4\sqrt{x-4}+4}\)

\(=\sqrt{\left(\sqrt{x-4}+2\right)^2}+\sqrt{\left(\sqrt{x-4}-2\right)^2}=\sqrt{x-4}+2+\left|\sqrt{x-4}-2\right|\)

Xét mẫu thức : \(\sqrt{\frac{16}{x^2}-\frac{8}{x}+1}=\sqrt{\left(\frac{4}{x}-1\right)^2}=\left|\frac{4}{x}-1\right|=\left|\frac{x-4}{x}\right|\)

Từ đó rút gọn P

bạn tìm giúp mình minP với

\(\left(\frac{1}{x}-\frac{2}{3}\right)^2-\frac{1}{16}=0\)

\(\Leftrightarrow\left(\frac{1}{x}-\frac{2}{3}\right)^2-\left(\frac{1}{4}\right)^2=0\)

\(\Leftrightarrow\left(\frac{1}{x}-\frac{2}{3}+\frac{1}{4}\right)\left(\frac{1}{x}-\frac{2}{3}-\frac{1}{4}\right)=0\)

\(\Leftrightarrow\left(\frac{1}{x}-\frac{5}{12}\right)\left(\frac{1}{x}-\frac{11}{12}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{1}{x}-\frac{5}{12}=0\\\frac{1}{x}-\frac{11}{12}=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{1}{x}=\frac{5}{12}\\\frac{1}{x}=\frac{11}{12}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{12}{11}\\x=\frac{12}{5}\end{cases}}\)

Vậy....

\(\left(\frac{1}{x}-\frac{2}{3}\right)^2-\frac{1}{16}=0\)

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

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

\(\Rightarrow\frac{1}{x}-\frac{2}{3}=\frac{1}{4}\)

\(\Rightarrow\frac{1}{x}=\frac{11}{12}\)

\(\Rightarrow x=\frac{11}{12}\)

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

=> \(\orbr{\begin{cases}\frac{1}{x}-\frac{2}{4}=\frac{1}{4}\\\frac{1}{x}-\frac{2}{4}=-\frac{1}{4}\end{cases}}\)

=>\(\orbr{\begin{cases}\frac{1}{x}=\frac{3}{4}\\\frac{1}{x}=\frac{1}{4}\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{4}{3}\\x=4\end{cases}}\)

học tốt

Theo đề ra ,ta có :

    - 1 / 12 < x < 1 / 8 mà x có giá trị nguyên 

=> x = 0

Ta có:\(\frac{4}{x}=\frac{2}{x+4}\)

\(\Leftrightarrow4x+16=2x\)

\(\Leftrightarrow4x-2x=-16\)

\(\Leftrightarrow2x=-16\)

\(\Leftrightarrow x=-8\)

hok tốt!!