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NV
Nguyễn Việt Lâm
Giáo viên
5 tháng 6 2019
ĐKXĐ:...
\(A=\left(\frac{\sqrt{a}+2}{\sqrt{a}\left(\sqrt{a}+2\right)}-\frac{\sqrt{a}-1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right).\frac{\sqrt{a}+1}{\sqrt{a}}=\left(\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{a}+1}\right).\frac{\left(\sqrt{a}+1\right)}{\sqrt{a}}\)
\(=\frac{1}{\sqrt{a}\left(\sqrt{a}+1\right)}.\frac{\left(\sqrt{a}+1\right)}{\sqrt{a}}=\frac{1}{a}\)
\(C=\left(\frac{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}{-\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\frac{\sqrt{x}\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right).\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{2\sqrt{x}-1}\)
\(=\left(\frac{\left(\sqrt{x}+1\right)}{-\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{x+\sqrt{x}+1}\right).\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(2\sqrt{x}-1\right)}.\frac{\left(2\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)}\)
\(=\left(-1+\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{x+\sqrt{x}+1}\right).\sqrt{x}=\left(\frac{-x-\sqrt{x}-1+x+\sqrt{x}}{x+\sqrt{x}+1}\right)\sqrt{x}=\frac{-\sqrt{x}}{x+\sqrt{x}+1}\)
FC
10 tháng 10 2019
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
22 tháng 5 2022
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
CW
25 tháng 6 2018
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
29 tháng 7 2020
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}
T
8 tháng 7 2019
a)ĐKXĐ: \(\left\{{}\begin{matrix}-2x+1\ne0\\\frac{3}{-2x+1}>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne\frac{1}{2}\\-2x+1>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne\frac{1}{2}\\x< \frac{1}{2}\end{matrix}\right.\)
b) ĐKXĐ: \(x-1\ge0\Leftrightarrow x\ge1\)
c) ĐKXĐ: \(x\in\mathbb{R}\)
ĐK: \(-1\le x\le1\)
Đặt \(\hept{\begin{cases}\sqrt{1+x}=a\\\sqrt{1-x}=b\end{cases}}\left(a,b\ge0\right)\)
\(\Rightarrow\hept{\begin{cases}a^2-b^2=2x\\a^2+b^2=2\end{cases}}\)
Từ đó ta có pt: \(\left(a-1\right)\left(b+1\right)=a^2-b^2\)
\(\Leftrightarrow ab+a-b-1=a^2-b^2\)
\(\Leftrightarrow2ab+2a-2b-2=2a^2-2b^2\)
\(\Leftrightarrow2ab+2a-2b-a^2-b^2=2a^2-2b^2\)
\(\Leftrightarrow a^2-2ab+b^2-2\left(a-b\right)+2a^2-2b^2=0\)
\(\Leftrightarrow\left(a-b\right)^2-2\left(a-b\right)+2\left(a-b\right)\left(a+b\right)=0\)
\(\Leftrightarrow\left(a-b\right)\left(a-b-2+2a+2b\right)=0\)
\(\Leftrightarrow\left(a-b\right)\left(3a+b-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=b\\3a+b-2=0\end{cases}}\)
Với a = b ta có: \(\sqrt{1+x}=\sqrt{1-x}\Rightarrow x=0\)
Với 3a + b - 2 = 0 ta có: \(3\sqrt{1+x}+\sqrt{1-x}-2=0\)
\(\Leftrightarrow3\sqrt{1+x}=2-\sqrt{1-x}\)
\(\Leftrightarrow9\left(1+x\right)=5-x-4\sqrt{1-x}\)
\(\Leftrightarrow-10x-4=4\sqrt{1-x}\)
\(\Rightarrow100x^2+80x+16=16-16x\)
\(\Rightarrow100x^2+96x=0\Rightarrow\orbr{\begin{cases}x=0\\x=-\frac{96}{100}\end{cases}}\) (tm)
Vậy pt có nghiệm \(x=0\) hoặc \(x=-\frac{96}{100}.\)