Sửa đề: \(C=\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}\)

a: ĐKXĐ: x>=1

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

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

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

\(x=3-\sqrt3\)

=>\(x-1=3-\sqrt3-1=2-\sqrt3=\frac12\left(4-2\sqrt3\right)\)

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

\(=\sqrt{\frac12\left(4-2\sqrt3\right)}+1+\left|\sqrt{\frac12\left(4-2\sqrt3\right)}-1\right|\)

\(=\frac{\sqrt3-1}{\sqrt2}+1+\left|\frac{\sqrt3-1}{\sqrt2}-1\right|=\frac{\sqrt3-1}{\sqrt2}+1+1-\frac{\sqrt3-1}{\sqrt2}=2\)

b: C=x-1

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

TH1: x>=2

=>x-1>=1

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

=>\(\sqrt{x-1}-1\ge0\)

(1) sẽ trở thành \(\sqrt{x-1}+1+\sqrt{x-1}-1=x-1\)

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

=>\(\sqrt{x-1}\left(\sqrt{x-1}-2\right)=0\)

=>\(\left[\begin{array}{l}x-1=0\\ x-1=4\end{array}\right.\Rightarrow\left[\begin{array}{l}x=1\left(loại\right)\\ x=5\left(nhận\right)\end{array}\right.\)

TH2: 1<=x<2

=>\(\sqrt{x-1}-1<0\)

(1) sẽ trở thành: \(\sqrt{x-1}+1-\sqrt{x-1}+1=x-1\)

=>x-1=2

=>x=3(loại)

a) Ta có: \(A=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}-\dfrac{\sqrt{x}}{3-\sqrt{x}}-\dfrac{3x+3}{x-9}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)

\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)

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

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

b) Ta có: \(x=\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}\)

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

=2

Thay x=2 vào A, ta được:

\(A=\dfrac{-3}{3+\sqrt{2}}=\dfrac{-9+3\sqrt{2}}{7}\)

ai h dung minh giai cho

a: Ta có: \(A=\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}-\dfrac{3\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)

\(=\dfrac{15\sqrt{x}-11-\left(3x+7\sqrt{x}-6\right)-\left(2x+\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}\)

a: Ta có: \(P=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)

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

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

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

$b$ là gì hả bạn?

Akai Haruma b là một số nào đó :))

a: \(B=\frac{2\sqrt{x}+13}{\left(\sqrt{x}+2\right)\cdot\left(\sqrt{x}+3\right)}+\frac{\sqrt{x}-2}{\sqrt{x}+2}\)

\(=\frac{2\sqrt{x}+13+\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{2\sqrt{x}+13+x+\sqrt{x}-6}{\left(\sqrt{x}+2\right)\cdot\left(\sqrt{x}+3\right)}=\frac{x+3\sqrt{x}+7}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)

b: C=A-B

\(=\frac{2\sqrt{x}-1}{\sqrt{x}+3}-\frac{x+3\sqrt{x}+7}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+2\right)-\left(x+3\sqrt{x}+7\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+2\right)}=\frac{2x+3\sqrt{x}-2-x-3\sqrt{x}-7}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+2\right)}\)

\(=\frac{x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+2\right)}=\frac{\sqrt{x}-3}{\sqrt{x}+2}\)

C nguyên khi \(\sqrt{x}-3\) ⋮\(\sqrt{x}+2\)

=>\(\sqrt{x}+2-5\) ⋮\(\sqrt{x}+2\)

=>-5⋮\(\sqrt{x}+2\)

=>\(\sqrt{x}+2=5\) (Do \(\sqrt{x}+2\ge2\forall x\) thỏa mãn ĐKXĐ)

=>\(\sqrt{x}=3\)

=>x=9(nhận)

Toán lớp 9 như này á hả

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