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Bài 1:
a) Ta có: \(\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\)
\(=\left(\sqrt{x}\right)^2-1^2\)
\(=x-1\)
b) Ta có: \(\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)\)
\(=\left(\sqrt{x}\right)^3+1^3\)
\(=x\sqrt{x}+1\)
c) Ta có: \(\left(2\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\)
\(=2x-2\sqrt{x}+\sqrt{x}-1\)
\(=2x-\sqrt{x}-1\)
Bài 2: Tìm x
a) Ta có: \(\sqrt{9x^2+6x+1}=3x-2\)
\(\Leftrightarrow\left|3x+1\right|=3x-2\)(*)
Trường hợp 1: \(x\ge\frac{-1}{3}\)
(*)\(\Leftrightarrow3x+1=3x-2\)
\(\Leftrightarrow3x+1-3x+2=0\)
\(\Leftrightarrow3=0\)(vô lý)
Trường hợp 2: \(x< \frac{-1}{3}\)
(*)\(\Leftrightarrow-3x-1=3x-2\)
\(\Leftrightarrow-3x-1-3x+2=0\)
\(\Leftrightarrow-6x+1=0\)
\(\Leftrightarrow-6x=-1\)
hay \(x=\frac{1}{6}\)(loại)
Vậy: \(S=\varnothing\)
b)Trường hợp 1: \(x\ge0\)
Ta có: \(\sqrt{x}-2>0\)
\(\Leftrightarrow\sqrt{x}>2\)
hay x>4(nhận)
Vậy: S={x|x>4}
Bài 2 xét x=0 => A =0
xét x>0 thì \(A=\frac{1}{x-2+\frac{2}{\sqrt{x}}}\)
để A nguyên thì \(x-2+\frac{2}{\sqrt{x}}\inƯ\left(1\right)\)
=>cho \(x-2+\frac{2}{\sqrt{x}}\)bằng 1 và -1 rồi giải ra =>x=?
1,Ta có \(\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2=a+b+c+2\sqrt{ab}+2\sqrt{bc}+2\sqrt{ac}\)
=> \(\sqrt{ab}+\sqrt{bc}+\sqrt{ac}=2\)
\(a+2=a+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}=\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)\)
\(b+2=\left(\sqrt{b}+\sqrt{c}\right)\left(\sqrt{b}+\sqrt{a}\right)\)
\(c+2=\left(\sqrt{c}+\sqrt{b}\right)\left(\sqrt{c}+\sqrt{a}\right)\)
=> \(\frac{\sqrt{a}}{a+2}+\frac{\sqrt{b}}{b+2}+\frac{\sqrt{c}}{c+2}=\frac{\sqrt{a}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)}+\frac{\sqrt{b}}{\left(\sqrt{b}+\sqrt{c}\right)\left(\sqrt{b}+\sqrt{a}\right)}+...\)
=> \(\frac{\sqrt{a}}{a+2}+...=\frac{2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)\left(\sqrt{b}+\sqrt{c}\right)}=\frac{4}{\sqrt{\left(a+2\right)\left(b+2\right)\left(c+2\right)}}\)
=> M=0
Vậy M=0
a)
Từ phương trình (2) ⇔ x = √2 - y√3 (3)
Thế (3) vào (1): ( √2 - y√3)√2 - y√3 = 1
⇔ √3y(√2 + 1) = 1 ⇔ y = =
Từ đó x = √2 - . √3 = 1.
Vậy có nghiệm (x; y) = (1; )
b)
Từ phương trình (2) ⇔ y = 1 - √10 - x√2 (3)
Thế (3) vào (1): x - 2√2(1 - √10 - x√2) = √5
⇔ 5x = 2√2 - 3√5 ⇔ x =
Từ đó y = 1 - √10 - . √2 =
Vậy hệ có nghiệm (x; y) = ;
c)
Từ phương trình (2) ⇔ x = 1 - (√2 + 1)y (3)
Thế (3) vào (1): (√2 - 1)[1 - (√2 + 1)y] - y = √2 ⇔ -2y = 1 ⇔ y = -
Từ đó x = 1 - (√2 + 1)(-) =
Vậy hệ có nghiệm (x; y) = (; -
)
b: \(=\dfrac{\left|x\right|+\left|x-2\right|+1}{2x-1}=\dfrac{x+x-2+1}{2x-1}=\dfrac{2x-1}{2x-1}=1\)
c: \(=\left|x-4\right|+\left|x-6\right|\)
=x-4+6-x=2
8)a) \(\left(x^2-9\right)\sqrt{2-x}=x\left(x^2-9\right)\)
\(\Leftrightarrow\left(x^2-9\right)\sqrt{2-x}-x\left(x^2-9\right)=0\)
\(\Leftrightarrow\left(x^2-9\right)\left(\sqrt{2-x}-x\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x\le2\\\left[{}\begin{matrix}x=\pm3\\\left\{{}\begin{matrix}x>0\\x^2+x-2=0\end{matrix}\right.\end{matrix}\right.\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\le2\\\left[{}\begin{matrix}x=\pm3\\\left\{{}\begin{matrix}x\ge0\\\left(x-1\right)\left(x+2\right)=0\end{matrix}\right.\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow x=-3\) hoặc x=1
Vậy nghiệm của pt là:...
a) đkxđ: \(\begin{cases}\sqrt{x^2-4}\ge0\\\sqrt{x^2}+4x+4\ge0\end{cases}\) \(\Leftrightarrow\begin{cases}\begin{cases}x-2\ge0\\x+2\ge0\end{cases}\\x+2\ge0\end{cases}\) \(\Leftrightarrow\begin{cases}x\ge2\\x\le-2\end{cases}\) \(\Leftrightarrow-2\ge x\ge2\)
\(\sqrt{x^2-4}+\sqrt{x^2+4x+4}=0\)
\(\Leftrightarrow\sqrt{\left(x-2\right)\left(x+2\right)}+\sqrt{\left(x+2\right)^2}=0\)
\(\Leftrightarrow\sqrt{\left(x-2\right)\left(x+2\right)}=x+2\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)=\left(x+2\right)^2\)
\(\Leftrightarrow\left(x+2\right)\left(x-2-x+2\right)=0\)
\(\Leftrightarrow x+2=0\)
\(\Leftrightarrow x=-2\)
S={-2}
b) đkxđ: \(\begin{cases}\sqrt{1-x^2}\ge0\\\sqrt{x+1}\ge0\end{cases}\) \(\Leftrightarrow\begin{cases}1-x^2\ge0\\x+1\ge0\end{cases}\) \(\Leftrightarrow\begin{cases}x^2\le1\\x\ge-1\end{cases}\) \(\Leftrightarrow\begin{cases}\begin{cases}x\le1\\x\ge-1\end{cases}\\x\ge-1\end{cases}\) \(\Leftrightarrow-1\le x\le1\)
\(\sqrt{1-x^2}+\sqrt{x+1}=0\)
\(\Leftrightarrow\sqrt{1-x^2}=-\sqrt{x+1}\)
\(\Leftrightarrow1-x^2=x+1\)
\(\Leftrightarrow-x-x^2=0\)
\(\Leftrightarrow-x\left(1+x\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}-x=0\\1+x=0\end{array}\right.\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\left(N\right)\\x=-1\left(N\right)\end{array}\right.\)
S={-1;0}
a) \(2\sqrt{2x}-5\sqrt{8x}+7\sqrt{18x}=28\) (*)
đk: x >/ 0
(*) \(\Leftrightarrow2\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}=28\)
\(\Leftrightarrow13\sqrt{2x}=28\) \(\Leftrightarrow\sqrt{2x}=\dfrac{28}{13}\Leftrightarrow2x=\left(\dfrac{28}{13}\right)^2\Leftrightarrow x=\dfrac{392}{169}\left(N\right)\)
Kl: \(x=\dfrac{392}{169}\)
b) \(\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4\) (*)
đk: x >/ 5
(*) \(\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
\(\Leftrightarrow2\sqrt{x-5}=4\Leftrightarrow\sqrt{x-5}=2\Leftrightarrow x-5=4\Leftrightarrow x=9\left(N\right)\)
Kl: x=9
c) \(\sqrt{\dfrac{3x-2}{x+1}}=2\) (*)
Đk: \(\left[{}\begin{matrix}x< -1\\x\ge\dfrac{2}{3}\end{matrix}\right.\)
(*) \(\Leftrightarrow\dfrac{3x-2}{x+1}=4\Leftrightarrow3x-2=4x+4\Leftrightarrow x=-6\left(N\right)\)
Kl: x=-6
d) \(\dfrac{\sqrt{5x-4}}{\sqrt{x+2}}=2\) (*)
Đk: \(x\ge\dfrac{4}{5}\)
(*) \(\Leftrightarrow\sqrt{5x-4}=2\sqrt{x+2}\Leftrightarrow5x-4=4x+8\Leftrightarrow x=12\left(N\right)\)
Kl: x=12
a, Điều kiện x ∉ {\(\frac{5}{3};\frac{1}{7}\)}
\(\sqrt{3x-5}=\sqrt{7x-1}\)
\(\left(\sqrt{3x-5}\right)^2=\left(\sqrt{7x-1}\right)^2\)
\(\left|3x-5\right|=\left|7x-1\right|\)
\(3x-5=7x-1\)
\(-4x=4\) => x = -1
a: \(x^2-2-x+\sqrt{2}=0\)
=>\(\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)-\left(x-\sqrt{2}\right)=0\)
=>\(\left(x-\sqrt{2}\right)\left(x+\sqrt{2}-1\right)=0\)
=>\(\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}+1\end{matrix}\right.\)
b: \(\left(1-\sqrt{2}\right)x^2-2\left(1+\sqrt{2}\right)x+1+3\sqrt{2}=0\)
\(\Delta=\left(-2-2\sqrt{2}\right)^2-4\left(1-\sqrt{2}\right)\left(1+3\sqrt{2}\right)\)
\(=12+8\sqrt{2}+4\left(\sqrt{2}-1\right)\left(3\sqrt{2}+1\right)\)
\(=12+8\sqrt{2}+4\left(6+\sqrt{2}-3\sqrt{2}-1\right)\)
\(=12+8\sqrt{2}+24-8\sqrt{2}-4=32>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{2\left(1+\sqrt{2}\right)-4\sqrt{2}}{2\left(1-\sqrt{2}\right)}=1\\x_2=\dfrac{2\left(1+\sqrt{2}\right)+4\sqrt{2}}{2\left(1-\sqrt{2}\right)}=-7-4\sqrt{2}\end{matrix}\right.\)