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VP
7 tháng 5 2018
a)\(\sqrt{4x}< =10\)
<=> 4x <= 100
<=> x <= 25
b) \(\sqrt{9x}>=3\)
<=> 9x >= 9
<=> x >= 1
c) \(\sqrt{4x^2+4x+1}=6\)
<=>\(\sqrt{\left(2x\right)^2+2\left(2x\right).1+1^2}=6\)
<=>\(\sqrt{\left(2x+1\right)^2}=6\)
<=>\(|2x+1|=6\)
<=>\(\orbr{\begin{cases}2x+1=6\\2x+1=-6\end{cases}}\)
<=>\(\orbr{\begin{cases}2x=5\\2x=-7\end{cases}}\)
<=>\(\orbr{\begin{cases}x=\frac{5}{2}\\x=\frac{-7}{2}\end{cases}}\)
d)\(\sqrt{9x-9}-2\sqrt{x-1}=6\)
<=>\(\sqrt{9\left(x-1\right)}-2\sqrt{x-1}=6\)
<=>\(3\sqrt{x-1}-2\sqrt{x-1}=6\)
<=>\(\sqrt{x-1}=6\)
<=> x - 1 = 36
<=> x = 37
f) \(\sqrt{2x+1}=\sqrt{x-1}\)
<=> 2x + 1 = x -1
<=> 2x - x = -1 -1
<=> x = -2
g)\(\sqrt{x^2-x-1}=\sqrt{x-1}\)
<=>x2 -x -1 = x -1
<=> x2 -x-x-1+1 = 0
<=> x2 - 2x + 0 = 0
<=> x(x-2) = 0
<=>\(\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)
<=>\(\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
YH
18 tháng 8 2019
a)...ghi lại đề...
\(\Leftrightarrow\sqrt{x^2-x-2x+2}=\sqrt{x-1}\)
\(\Leftrightarrow\sqrt{x\left(x-1\right)-2\left(x-1\right)}=\sqrt{x-1}\)
\(\Leftrightarrow\sqrt{\left(x-2\right)\left(x-1\right)}=\sqrt{x-1}\)
\(\Leftrightarrow\sqrt{x-2}\cdot\sqrt{x-1}=\sqrt{x-1}\)
\(\Leftrightarrow\sqrt{x-2}=\frac{\sqrt{x-1}}{\sqrt{x-1}}=1\)
\(\Leftrightarrow\sqrt{x-2}^2=1^2\)
\(\Leftrightarrow x-2=1\)(Vì \(x-2\ge0\Leftrightarrow x\ge2\))
\(\Leftrightarrow x=3\)
\(\)
18 tháng 8 2019
\(a,\sqrt{x^2-3x+2}=\sqrt{x-1}\)
\(\Rightarrow x^2-3x+2=x-1\)
\(\Rightarrow x^2-4x+3=0\)
\(\Rightarrow x^2-x-3x+3=0\)
\(\Rightarrow\left(x-3\right)\left(x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}}\)
Vậy..........
NV
Nguyễn Việt Lâm
Giáo viên
25 tháng 5 2019
\(\sqrt{4x^2}=6\Rightarrow\left|2x\right|=6\Rightarrow\left[{}\begin{matrix}2x=6\\2x=-6\end{matrix}\right.\) \(\Rightarrow x=\pm3\)
b/ ĐKXĐ: \(x\ge0\)
\(\sqrt{16x}=8\Leftrightarrow16x=64\Rightarrow x=4\)
c/ ĐKXĐ: \(x\ge1\)
\(\sqrt{9\left(x-1\right)}=21\Leftrightarrow\sqrt{x-1}=7\Leftrightarrow x-1=49\Rightarrow x=50\)
d/ \(\sqrt{4\left(1-x\right)^2}=6\Leftrightarrow2\left|1-x\right|=6\Leftrightarrow\left|1-x\right|=3\Rightarrow\left[{}\begin{matrix}x=-2\\x=4\end{matrix}\right.\)
e/ \(\sqrt{1-4x+4x^2}=5\Leftrightarrow\sqrt{\left(2x-1\right)^2}=5\Leftrightarrow\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
f/ĐKXĐ: \(x\ge-\frac{1}{2}\)
\(\sqrt{9x^2}=2x+1\Leftrightarrow\left|3x\right|=2x+1\Leftrightarrow\left[{}\begin{matrix}3x=2x+1\\-3x=2x+1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=-\frac{1}{5}\end{matrix}\right.\)
AH
Akai Haruma
Giáo viên
19 tháng 8 2019
a)
ĐK: $x\geq 2$
PT \(\Leftrightarrow \sqrt{(x-1)(x-2)}=\sqrt{x-1}\)
\(\Leftrightarrow \sqrt{x-1}(\sqrt{x-2}-1)=0\)
\(\Rightarrow \left[\begin{matrix} \sqrt{x-1}=0\\ \sqrt{x-2}-1=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=1(\text{loại vì x}\geq 2)\\ \sqrt{x-2}=1\end{matrix}\right.\)
\(\Rightarrow x=1^2+2=3\) là nghiệm duy nhất thỏa mãn
b)
ĐK: $x\in\mathbb{R}$
Bình phương 2 vế:
\(\Rightarrow x^2-4x+4=4x^2-12x+9\)
\(\Leftrightarrow (x-2)^2=(2x-3)^2\)
\(\Leftrightarrow (x-2)^2-(2x-3)^2=0\Leftrightarrow (x-2-2x+3)(x-2+2x-3)=0\)
\(\Leftrightarrow (-x+1)(3x-5)=0\Rightarrow \left[\begin{matrix} x=1\\ x=\frac{5}{3}\end{matrix}\right.\) (đều thỏa mãn)
Vậy..........
AH
Akai Haruma
Giáo viên
19 tháng 8 2019
c)
ĐKXĐ: $x\geq 3$
PT \(\Leftrightarrow \sqrt{(x-2)(x-3)}=\sqrt{x-2}\)
\(\Leftrightarrow (x-2)(x-3)=x-2\) (bình phương 2 vế không âm)
\(\Leftrightarrow (x-2)(x-3-1)=0\)
\(\Rightarrow \left[\begin{matrix} x-2=0\\ x-4=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=2(\text{loại vì x}\geq 3)\\ x=4\end{matrix}\right.\)
Vậy $x=4$
d)
ĐK: $x\in\mathbb{R}$
PT \(\Leftrightarrow 4x^2-4x+1=x^2-6x+9\) (bình phương 2 vế không âm)
\(\Leftrightarrow (2x-1)^2=(x-3)^2\Leftrightarrow (2x-1)^2-(x-3)^2=0\)
\(\Leftrightarrow (2x-1-x+3)(2x-1+x-3)=0\)
\(\Leftrightarrow (x+2)(3x-4)=0\Rightarrow \left[\begin{matrix} x+2=0\\ 3x-4=0\end{matrix}\right.\)
\(\Leftrightarrow \left[\begin{matrix} x=-2\\ x=\frac{4}{3}\end{matrix}\right.\) (đều thỏa mãn)
Vậy.........
1 tháng 8 2020
Bài 1: Giải phương trình
a) ĐKXĐ: \(x\ge3\)
Ta có: \(\sqrt{100\cdot\left(x-3\right)}=\sqrt{20}\)
\(\Leftrightarrow\left|100\cdot\left(x-3\right)\right|=\left|20\right|\)
\(\Leftrightarrow100\cdot\left|x-3\right|=20\)
\(\Leftrightarrow\left|x-3\right|=\frac{1}{5}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=\frac{1}{5}\\x-3=-\frac{1}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{16}{5}\left(nhận\right)\\x=\frac{14}{5}\left(loại\right)\end{matrix}\right.\)
Vậy: \(S=\left\{\frac{16}{5}\right\}\)
b) Ta có: \(\sqrt{\left(x-3\right)^2}=7\)
\(\Leftrightarrow\left|x-3\right|=7\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=7\\x-3=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-4\end{matrix}\right.\)
Vậy: S={10;-4}
c) Ta có: \(\sqrt{4x^2+4x+1}=6\)
\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)
\(\Leftrightarrow\left|2x+1\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{5}{2}\\x=\frac{-7}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{\frac{5}{2};\frac{-7}{2}\right\}\)
PA
31 tháng 7 2017
\(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}=16\)
\(\Leftrightarrow\sqrt{x+1}=4\)
<=> x + 1 = 16
<=> x = 15 (nhận)
~ ~ ~
\(\sqrt{4x+20}-3\sqrt{5+x}+\dfrac{4}{3}\sqrt{9x+45}=6\)
\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)
\(\Leftrightarrow3\sqrt{x+5}=6\)
\(\Leftrightarrow\sqrt{x+5}=2\)
<=> x + 5 = 4
<=> x = - 1 (nhận)
MY
1
G
24 tháng 6 2019
a) ĐKXĐ: \(x\ge-4\)
a) Ta có: \(\sqrt{6-4x+x^2}=x+4\Rightarrow\left(x+4\right)^2=x^2-4x+6\)
\(\Rightarrow x^2+8x+16=x^2-4x+6\Rightarrow4x+10=0\Rightarrow x=-\frac{5}{2}\left(loại\right)\)
Vậy pt vô nghiệm
b) \(\sqrt{4x^2-4x+1}+\sqrt{2x-1}=0\Rightarrow\sqrt{\left(2x-1\right)^2}+\sqrt{2x-1}=0\)
\(\Leftrightarrow\sqrt{2x-1}\left(\sqrt{2x-1}+1\right)=0\Rightarrow x=\frac{1}{2}\)
25 tháng 6 2018
\(A=2\left|x\right|-4x+1\) \(\forall x\ge0\) A=\(2x-4x+1=1-2x\)
\(\forall x< 0\) A=\(-2x-4x+1=1-6x\)
B=\(\sqrt{x-1-2\sqrt{x-1}+1}=\sqrt{\left(x-1-1\right)^2}=\sqrt{\left(x-2\right)^2}=\left|x-2\right|\)
\(\forall x\ge2\) B = x-2 \(\forall x< 2\) B = 2-x
C=\(\sqrt{x-3+2.3\sqrt{x-3}+9}=\sqrt{\left(x-3+3\right)^2}=\left|x\right|\)
\(\forall x\ge0\) C=x \(\forall x< 0\) C=-x
25 tháng 6 2018
\(a.A=\sqrt{4x^2}-4x+1=|2x|-4x+1\)
\(b.B=\sqrt{x-2\sqrt{x-1}}=\sqrt{x-1-2\sqrt{x-1}+1}=\sqrt{\left(\sqrt{x-1}-1\right)^2}=|\sqrt{x-1}+1|=\sqrt{x-1}+1\)
\(c.C=\sqrt{x+6+6\sqrt{x-3}}=\sqrt{x-3+6\sqrt{x-3}+9}=\sqrt{\left(\sqrt{x-3}+3\right)^2}=|\sqrt{x-3}+3|=\sqrt{x-3}+3\left(x\ge3\right)\)
\(d.D=\sqrt{x+2}+\dfrac{1}{\sqrt{x^2+2x+1}}=\sqrt{x+2}+\dfrac{1}{\sqrt{\left(x+1\right)^2}}=\sqrt{x+2}+\dfrac{1}{|x+1|}=\sqrt{x+2}+\dfrac{1}{x+1}\left(x\ge-2\right)\)
Ta có:
\(\sqrt{4x^2+4x+1}=6\)
\(\Leftrightarrow4x^2+4x+1=36\)
\(\Leftrightarrow4x\left(x+1\right)+1=36\)
\(\Leftrightarrow4x\left(x+1\right)=35\)
\(\sqrt{4x^2+4x+1}=6\)
\(\sqrt{\left(2x+1\right)^2}=6\)
\(\Rightarrow/2x+1/=6\)
TH1 : \(2x+1\ge0\Rightarrow2x\ge-1\Rightarrow x< \frac{-1}{2}\)
\(\Rightarrow2x+1=6\)
\(\Rightarrow2x=5\)\(\Rightarrow x=\frac{5}{2}\)
TH2 : \(2x+1< 0\Rightarrow2x=-1\Rightarrow x>\frac{-1}{2}\)
\(\Rightarrow2x+1=-6\)
\(\Rightarrow2x=-7\Rightarrow x=\frac{-7}{2}\)