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

\(=\sqrt{x}+2-3+\sqrt{x}=2\sqrt{x}-1\)

ĐK: \(x>0\)

Khi đó:

\(P=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}}-3+\sqrt{x}\\ =\sqrt{x}+2-3+\sqrt{x}\\ =2\sqrt{x}-1\)

ĐKXĐ: x>=1/2

Ta có: \(P=\sqrt{x+\sqrt{2x-1}}-\sqrt{x-\sqrt{2x-1}}\)

\(=\frac12\left(\sqrt{2x+2\sqrt{2x-1}}-\sqrt{2x-2\sqrt{2x-1}}\right)\)

\(=\frac12\left(\sqrt{2x-1+2\cdot\sqrt{2x-1}\cdot1+1}+\sqrt{2x-1-2\cdot\sqrt{2x-1}\cdot1+1}\right)\)

\(=\frac12\left(\sqrt{\left(\sqrt{2x-1}+1\right)^2}+\sqrt{\left(\sqrt{2x-1}-1\right)^2}\right)\)

\(=\frac12\left(\sqrt{2x-1}+1+\left|\sqrt{2x-1}-1\right|\right)\)

TH1: \(\sqrt{2x-1}-1\ge0\)

=>\(\sqrt{2x-1}\ge1\)

=>2x-1>=1

=>2x>=2

=>x>=1

\(A=\frac12\left(\sqrt{2x-1}+1+\left|\sqrt{2x-1}-1\right|\right)\)

\(=\frac12\left(\sqrt{2x-1}+1+\sqrt{2x-1}-1\right)=\frac12\cdot2\sqrt{2x-1}=\sqrt{2x-1}\)

TH2: 1/2<=x<1

\(A=\frac12\left(\sqrt{2x-1}+1+\left|\sqrt{2x-1}-1\right|\right)\)

\(=\frac12\left(\sqrt{2x-1}+1+1-\sqrt{2x-1}\right)\)

\(=\frac12\cdot2=1\)

Alo đề nghị viết đề một cách chính xác 

1: Ta có: \(x^2-2x+5-\left(x-7\right)\left(x+2\right)\)

\(=x^2-2x+5-x^2-2x+7x-14\)

\(=3x-9\)

2: Ta có: \(-5x\left(x-5\right)+\left(x-3\right)\left(x^2-7\right)\)

\(=-5x^2+25x+x^3-7x-3x^2+21\)

\(=x^3-8x^2+18x+21\)

3: Ta có: \(x\left(x^2-x-2\right)-\left(x+5\right)\left(x-1\right)\)

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

\(=x^3-2x^2-6x+5\)

1, 99 x 1 = 99

2, 88 x 2 = 176

3, 7 x 9 x 3 x 6 + 2 = 1136

4, 3x + 9 = 0

3x = 0 - 9

3x = -9

x = ( -9 ) : 3

x = -3

Vậy x = -3

~ Hok tốt ~

1.99

2.176

3.1136

4.x=3

1/3(2x-5)=-2/3-3/2=-13/6

2x-5=-13/6:1/3=-13/2

2x=-13/2+5

2x=-3/2

x=-3/2:2=-3/4

       Trương Ứng Hòa

Conan bn sai rùi nhé tại sao chỉ còn 1/3 ? còn (-1/3)chứ

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

b: Để M=7 thì (x+3)(x-2)=0

=>x=-3(nhận) hoặc x=2(loại)

Vậy: x=-3