a: \(\left(1-\sqrt{x}\right)\left(1+\sqrt{x}+x\right)-\sqrt{x^3}\)

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

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

b: \(\left(\dfrac{1-\sqrt{a}}{1-a}\right)^2\cdot\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\)

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

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

\(=\dfrac{\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)^2}=1\)

Bạn nên viết đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để được hỗ trợ tốt hơn.

a: \(x\ge1\frac15\)

=>\(x-1\frac15>0;x-\frac15>0\)

=>\(M=x-1\frac15+x-\frac15-3\frac15=2x-\frac65-\frac15-\frac{16}{5}=2x-\frac{23}{5}\)

b: \(x\le\frac15\)

=>\(x-\frac15\le0\)

=>\(x-\frac15-1\le-1<0\)

=>\(x-1\frac15<0\)

Ta có: \(M=\left|1\frac15-x\right|+\left|x-\frac15\right|-3\frac15\)

\(=\frac65-x+\frac15-x-\frac{16}{5}=\frac{-9}{5}-2x\)

c: \(\frac15

=>\(x-\frac15>0;x-\frac65<0\)

Ta có: \(M=\left|1\frac15-x\right|+\left|x-\frac15\right|-3\frac15\)

\(=\frac65-x+x-\frac15-3-\frac15=1-3-\frac15=-2-\frac15=-\frac{11}{5}\)

a: \(x\ge1\frac15\)

=>\(x-1\frac15>0;x-\frac15>0\)

=>\(M=x-1\frac15+x-\frac15-3\frac15=2x-\frac65-\frac15-\frac{16}{5}=2x-\frac{23}{5}\)

b: \(x\le\frac15\)

=>\(x-\frac15\le0\)

=>\(x-\frac15-1\le-1<0\)

=>\(x-1\frac15<0\)

Ta có: \(M=\left|1\frac15-x\right|+\left|x-\frac15\right|-3\frac15\)

\(=\frac65-x+\frac15-x-\frac{16}{5}=\frac{-9}{5}-2x\)

c: \(\frac15

=>\(x-\frac15>0;x-\frac65<0\)

Ta có: \(M=\left|1\frac15-x\right|+\left|x-\frac15\right|-3\frac15\)

\(=\frac65-x+x-\frac15-3-\frac15=1-3-\frac15=-2-\frac15=-\frac{11}{5}\)

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

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

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

a)      Cách 1:

\(6(y - x) - 2(x - y)\)

\( = 6y - 6x - 2x + 2y\)

\( = 8y - 8x\)

Cách 2:

\(6(y - x) - 2(x - y)\\= 6(y-x)+2(y-x)\\=(6+2).(y-x)\\=8.(y-x)\\=8y-8x\)

b)      \(3{x^2} + x - 4x - 5{x^2}\)

\( = (3{x^2} - 5{x^2}) + (x - 4x)\)

\( =  - 2{x^2} - 3x\)

a: \(x\ge1\frac15\)

=>\(x-1\frac15>0;x-\frac15>0\)

=>\(M=x-1\frac15+x-\frac15-3\frac15=2x-\frac65-\frac15-\frac{16}{5}=2x-\frac{23}{5}\)

b: \(x\le\frac15\)

=>\(x-\frac15\le0\)

=>\(x-\frac15-1\le-1<0\)

=>\(x-1\frac15<0\)

Ta có: \(M=\left|1\frac15-x\right|+\left|x-\frac15\right|-3\frac15\)

\(=\frac65-x+\frac15-x-\frac{16}{5}=\frac{-9}{5}-2x\)

c: \(\frac15

=>\(x-\frac15>0;x-\frac65<0\)

Ta có: \(M=\left|1\frac15-x\right|+\left|x-\frac15\right|-3\frac15\)

\(=\frac65-x+x-\frac15-3-\frac15=1-3-\frac15=-2-\frac15=-\frac{11}{5}\)