Bo thi:>

+ đk x > 0 , x khác 1

ĐKXĐ: \(x\ge-2\)

- Với \(-2\le x< 0\Rightarrow\left\{{}\begin{matrix}\sqrt{x^2+1}>1\Rightarrow\sqrt{x^2+1}-x>1\\\sqrt{x+3}\ge1\Rightarrow\sqrt{x+2}+\sqrt{x+3}\ge1\end{matrix}\right.\)

\(\Rightarrow\left(\sqrt{x^2+1}-x\right)\left(\sqrt{x+2}+\sqrt{x+3}\right)>1\) pt vô nghiệm

- Với \(x\ge0\)

\(\Leftrightarrow\frac{1}{\sqrt{x^2+1}+x}\left(\sqrt{x+2}+\sqrt{x+3}\right)=1\)

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

\(\Leftrightarrow\sqrt{x^2+1}-\sqrt{x+3}+x-\sqrt{x+2}=0\)

\(\Leftrightarrow\frac{x^2-x-2}{\sqrt{x^2+1}+\sqrt{x+3}}+\frac{x^2-x-2}{x+\sqrt{x+2}}=0\)

\(\Leftrightarrow\left(x^2-x-2\right)\left(\frac{1}{\sqrt{x+2}+\sqrt{x+3}}+\frac{1}{x+\sqrt{x+2}}\right)=0\)

\(\Leftrightarrow x^2-x-2=0\Leftrightarrow x=2\)

Vậy pt có nghiệm duy nhất \(x=2\)

Mọi người ơi, giúp em với ạ!

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

a/

Đặt \(\left\{{}\begin{matrix}\sqrt[3]{x-1}=a\\\sqrt[3]{27-14x}=b\end{matrix}\right.\) ta được hệ:

\(\left\{{}\begin{matrix}2a+b=1\\14a^3+b^3=13\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}b=1-2a\\14a^3+b^3=13\end{matrix}\right.\)

\(\Rightarrow14a^3+\left(1-2a\right)^3=13\)

\(\Leftrightarrow a^3+2a^2-a-2=0\)

\(\Leftrightarrow\left(a-1\right)\left(a+1\right)\left(a+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=1\\x-1=-1\\x-1=-8\end{matrix}\right.\) \(\Leftrightarrow...\)

b/ ĐKXĐ: ...

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

\(VP=\left(x-3\right)^2+2\ge2\)

Đẳng thức xảy ra khi và chỉ khi \(x=3\)

1) Ta có: \(\sqrt{21-x}+1=x\)

\(\Leftrightarrow21-x=\left(x-1\right)^2\)

\(\Leftrightarrow x^2-2x+1-21+x=0\)

\(\Leftrightarrow x^2-3x-20=0\)

\(\text{Δ}=\left(-3\right)^2-4\cdot1\cdot\left(-20\right)=9+80=89\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{3+\sqrt{89}}{2}\\x_2=\dfrac{3-\sqrt{89}}{2}\end{matrix}\right.\)

1)\(\sqrt{21-x}+1=x\)

\(\Leftrightarrow21-x=\left(x-1\right)^2\)

\(\Leftrightarrow21-x=x^2-2x+1\)

\(\Leftrightarrow x^2-x-20=0\)

\(\Leftrightarrow\left(x-5\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-4\end{matrix}\right.\)

2)\(\sqrt{8-x}+2=x\)

\(\Leftrightarrow8-x=\left(x-2\right)^2\)

\(\Leftrightarrow8-x=x^2-4x+4\)

\(\Leftrightarrow x^2-3x-4=0\Leftrightarrow\left(x-4\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-1\end{matrix}\right.\)

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

\(\Leftrightarrow3x+1=\left(3x-1\right)^2\)

\(\Leftrightarrow3x+1=9x^2-6x+1\)

\(\Leftrightarrow9x^2-9x=0\Leftrightarrow9x\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

\(a,A=\left(\dfrac{x+14\sqrt{x}-5}{x-25}+\dfrac{\sqrt{x}}{\sqrt{x}+5}\right):\dfrac{\sqrt{x}+2}{\sqrt{x}-5}\)

\(\Rightarrow A=\left(\dfrac{x+14\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\right).\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)

\(\Rightarrow A=\left(\dfrac{x+14\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}+\dfrac{x-5\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\right).\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)

\(\Rightarrow A=\dfrac{x+14\sqrt{x}-5+x-5\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}.\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)

\(\Rightarrow A=\dfrac{2x+9\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}.\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)

\(\Rightarrow A=\dfrac{2x+10\sqrt{x}-\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+2\right)}\)

\(\Rightarrow A=\dfrac{2\sqrt{x}\left(\sqrt{x}+5\right)-\left(\sqrt{x}+5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+2\right)}\)

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

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

ĐKXĐ : \(x\ge2\)

Ta có : \(A=\dfrac{x+3\sqrt{x-2}}{x+4\sqrt{x-2}+1}\) . Đặt t = \(\sqrt{x-2}\ge0\) \(\Rightarrow x=t^2+2\)

Khi đó : \(A=\dfrac{t^2+2+3t}{t^2+4t+3}=\dfrac{\left(t+2\right)\left(t+1\right)}{\left(t+3\right)\left(t+1\right)}=\dfrac{t+2}{t+3}=1-\dfrac{1}{t+3}\ge1-\dfrac{1}{3}=\dfrac{2}{3}\)

" = " \(\Leftrightarrow t=0\Leftrightarrow x=2\)

Vậy ...