17/13 ; 14/23 ; 3/4; 9/13; 9/23; 1/4

17/13= 17/13 ,  14/23= 14/23 ,  75/100=3/4  ,   9/13=9/13  ,  9/23=9/23   ,  15/60=5/12

75/100 =15/20    15/60 = 3/12    còn lại giữ nguyên nha

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

rút gọn thì cứ tách ra đi bạn

mình làm rồi nhưng khong chắc là rút gọn được nữa không thôi

Lời giải:

\(\frac{2x-2\sqrt{x}+2}{x-\sqrt{x}}=\frac{2(x-\sqrt{x})+2}{x-\sqrt{x}}=\frac{2(x-\sqrt{x})+2}{x-\sqrt{x}}=2+\frac{2}{x-\sqrt{x}}\)

\(\dfrac{2x-2\sqrt{x}+2}{x\sqrt{x}+1}=\dfrac{2}{\sqrt{x}+1}\)

Với `x >= 0,x \ne 1` có:

`C=A/B=A:B=[\sqrt{x}+1]/[x+\sqrt{x}+1]:(\sqrt{x}/[x\sqrt{x}-1]+1/[\sqrt{x}-1])`

`C=[\sqrt{x}+1]/[x+\sqrt{x}+1]:[\sqrt{x}+x+\sqrt{x}+1]/[(\sqrt{x}-1)(x+\sqrt{x}+1)]`

`C=[\sqrt{x}+1]/[x+\sqrt{x}+1].[(\sqrt{x}-1)(x+\sqrt{x}+1)]/[x+2\sqrt{x}+1]`

`C=[\sqrt{x}+1]/[x+\sqrt{x}+1].[(\sqrt{x}-1)(x+\sqrt{x}+1)]/[(\sqrt{x}+1)^2]`

`C=[\sqrt{x}-1]/[\sqrt{x}+1]`

1.Thế \(x=4\) vào A, ta được:

\(A=\dfrac{\sqrt{4}+1}{4+\sqrt{4}+1}=\dfrac{2+1}{4+2+1}=\dfrac{3}{7}\)

2.

\(B=\dfrac{\sqrt{x}}{x\sqrt{x}-1}+\dfrac{1}{\sqrt{x}-1}\)

\(B=\dfrac{\sqrt{x}}{\sqrt{x}^3-1}+\dfrac{1}{\sqrt{x}-1}\)

\(B=\dfrac{\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{1}{\sqrt{x}-1}\)

\(B=\dfrac{\sqrt{x}+\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)

\(B=\dfrac{x+2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)

\(B=\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)

\(C=\dfrac{A}{B}\)

\(C=\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}:\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)

\(C=\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}.\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2}\)

\(C=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)

\(C=A:B=\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}:\left(\dfrac{\sqrt{x}+x+\sqrt{x}+1}{x\sqrt{x}-1}\right)\)

\(=\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\cdot\dfrac{x\sqrt{x}-1}{\left(\sqrt{x}+1\right)^2}\)

\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)

ĐKXĐ: \(x\ge0;x\ne4\)

\(A=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\dfrac{3\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\dfrac{12}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{x+4\sqrt{x}+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\dfrac{3\sqrt{x}-6}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+2\right)}-\dfrac{12}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{x+4\sqrt{x}+4-3\sqrt{x}+6-12}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{x+\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}\)

\(C=\left(\dfrac{3}{x-1}+\dfrac{1}{\sqrt{x}+1}\right):\dfrac{1}{\sqrt{x}+1}\)

\(=\dfrac{3+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{1}\)

\(=\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\)

a: 7A-2B

\(=7\cdot\left(5x+2y\right)-2\left(9x+7y\right)\)

\(=35x+14y-18x-14y=17x\)

b: \(7\left(5x+2y\right)+2\left(9x+7y\right)=17y⋮17\)

mà \(5x+2y⋮17\)

nên \(2\left(9x+7y\right)⋮17\)

=>\(9x+7y⋮17\)

Gấp lắm ạ!

\(\dfrac{25.12+25.17}{29.\left(-13\right)+29.\left(-27\right)}=\dfrac{25.\left(12+17\right)}{29.\left[\left(-13\right)+\left(-17\right)\right]}=\dfrac{25.29}{29.\left(-30\right)}=\dfrac{725}{-870}\)

\(\dfrac{725}{\text{-870 }}\)