B

Chọn B

`C4:`

`x/[x-2]+2/[2-x]=x/[x-2]-2/[x-2]=[x-2]/[x-2]=1`

`C5:[15x^2]/[17y^4] . [34y^5]/[15x^3]`

 `=[15x^2 . 2.17y^4 . y]/[17y^4 . 15x^2 . x]=[2y]/x`

`C5:[6x+18]/[(x+4)^2]:[3(x+3)]/[x+4]`

 `=[6(x+3)]/[(x+4)^2] . [x+4]/[3(x+3)]`

`=2/[x+4]`

a: |x-1|=3

=>x-1=3 hoặc x-1=-3

=>x=-2(nhận) hoặc x=4(loại)

Khi x=-2 thì \(A=\dfrac{4+4}{-2-4}=\dfrac{8}{-6}=\dfrac{-4}{3}\)

b: ĐKXĐ: x<>4; x<>-4

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

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

=-4x/x-4

c: A+B

=-4x/x-4+x^2+4/x-4

=(x-2)^2/(x-4)
A+B>0

=>x-4>0

=>x>4

\(=2x^2\cdot\dfrac{3}{x^2}=6\)

\(a,A=4\sqrt{3}-5\sqrt{3}+2-\sqrt{3}=2-2\sqrt{3}\\ B=\dfrac{x+2\sqrt{x}+8+2\sqrt{x}-8}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+4\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-4}\\ b,B-\dfrac{1}{2}A=\dfrac{\sqrt{x}}{\sqrt{x}-4}-\dfrac{1}{2}\left(2-2\sqrt{3}\right)=0\\ \Leftrightarrow\dfrac{\sqrt{x}}{\sqrt{x}-4}=1+\sqrt{3}\\ \Leftrightarrow\sqrt{x}=\left(1+\sqrt{3}\right)\left(\sqrt{x}-4\right)\Leftrightarrow\sqrt{x}=\sqrt{x}-4\sqrt{3}+\sqrt{3x}-4\\ \Leftrightarrow\sqrt{3x}=4\sqrt{3}+4\\ \Leftrightarrow\sqrt{x}=\dfrac{4\sqrt{3}+4}{\sqrt{3}}\\ \Leftrightarrow\sqrt{x}=\dfrac{12+4\sqrt{3}}{3}\\ \Leftrightarrow x=\dfrac{192+96\sqrt{3}}{9}=\dfrac{64+32\sqrt{3}}{3}\)

\(\dfrac{\sqrt{x}}{\sqrt{x}-4}=1-\sqrt{3}\)
Nhỉ???

a: Ta có: \(x=\sqrt{28-16\sqrt{3}}+2\sqrt{3}\)

\(=4-2\sqrt{3}+2\sqrt{3}\)

=4

Thay x=4 vào B, ta được:

\(B=\dfrac{2-4}{2}=-1\)

Bạn nên gõ đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để mọi người hiểu đề và hỗ trợ bạn tốt hơn nhé.

\(A=\frac{y}{x}\cdot\sqrt{\frac{x^2}{y^4}}=\frac{y}{x}\cdot\frac{\sqrt{x^2}}{\sqrt{y^4}}=\frac{y}{x}\cdot\frac{\left|x\right|}{\left|y^2\right|}=\frac{y}{x}\cdot\frac{x}{y^2}=\frac{1}{y}\)( x > 0 ; y > 0 )