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3 tháng 7

\(-5\left(4x-1\right)\left(x-2\right)=2\left(4x-1\right)^2\\ \Leftrightarrow\left(4x-1\right)\left(-5x+10\right)-2\left(4x-1\right)^2=0\\ \Leftrightarrow\left(4x-1\right)\left(-5x+10-8x+2\right)=0\\ \Leftrightarrow\left(4x-1\right)\left(-13x+12\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}4x-1=0\\-13x+12=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=1\\13x=12\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=\dfrac{12}{13}\end{matrix}\right.\)

27 tháng 6 2023

a) \(\sqrt{1-4x+4x^2}=5\) 

\(\Leftrightarrow\sqrt{\left(1-2x\right)^2}=5\)

\(\Leftrightarrow\left|1-2x\right|=5\)

\(\Leftrightarrow2x-1=5\)

\(\Leftrightarrow2x=6\)

\(\Leftrightarrow x=3\)

b) \(\sqrt{x^2+6x+9}=3x-1\)

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

\(\Leftrightarrow\left|x+3\right|=3x-1\)

\(\Leftrightarrow x+3=3x-1\)

\(\Leftrightarrow2x=4\)

\(\Leftrightarrow x=2\)

27 tháng 6 2023

\(a,\sqrt{1-4x+4x^2}=5\\ \Leftrightarrow\sqrt{\left(1-2x\right)^2}=5\\ \Leftrightarrow\left|1-2x\right|=5\)

\(TH_1:x\le\dfrac{1}{2}\)

\(1-2x=5\\ \Leftrightarrow x=-2\left(tm\right)\)

\(TH_2:x\ge\dfrac{1}{2}\)

\(-1+2x=5\\ \Leftrightarrow x=3\left(tm\right)\)

Vậy \(S=\left\{-2;3\right\}\)

\(b,\sqrt{x^2+6x+9}=3x-1\\ \Leftrightarrow\sqrt{\left(x+3\right)^2}=3x-1\\ \Leftrightarrow\left|x+3\right|=3x-1\)

\(TH_1:x\ge-3\\ x+3=3x-1\\ \Leftrightarrow-2x=-4\Leftrightarrow x=2\left(tm\right)\)

\(TH_2:x< 3\\ -x-3=3x-1\\ \Leftrightarrow-4x=2\\ \Leftrightarrow x=-\dfrac{1}{2}\left(tm\right)\)

Vậy \(S=\left\{2;-\dfrac{1}{2}\right\}\)

25 tháng 10 2021

ĐKXĐ: \(x\ge1\)

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

\(\Leftrightarrow2x-1=x-1\left(do.x\ge1\right)\)

\(\Leftrightarrow x=0\left(ktm\right)\)

Vậy \(S=\varnothing\)

25 tháng 10 2021

ĐK \(x\ge1\)

\(\Leftrightarrow\left|2x-1\right|=x-1\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=x-1\\2x-1=1-x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=\dfrac{2}{3}\left(ktm\right)\end{matrix}\right.\)

AH
Akai Haruma
Giáo viên
28 tháng 10 2018

a) ĐK: \(x\geq 0\)

Ta có: \(\sqrt{x}+\sqrt{x+1}=1\Leftrightarrow \sqrt{x}+\sqrt{x+1}-1=0\)

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

\(\Leftrightarrow \sqrt{x}+\frac{x}{\sqrt{x+1}+1}=0\)

\(\Leftrightarrow \sqrt{x}\left(1+\frac{\sqrt{x}}{\sqrt{x+1}+1}\right)=0\)

Thấy rằng \(1+\frac{\sqrt{x}}{\sqrt{x+1}+1}>0, \forall x\geq 0\Rightarrow 1+\frac{\sqrt{x}}{\sqrt{x+1}+1}\neq 0\)

Do đó \(\sqrt{x}=0\Rightarrow x=0\) (thỏa mãn)

AH
Akai Haruma
Giáo viên
28 tháng 10 2018

b) ĐK: \(x\geq 1\)

Ta thấy với mọi \(x\geq 1\) thì:\(\left\{\begin{matrix} \sqrt{x+4}\geq \sqrt{1+4}>2 \\ \sqrt{x-1}\geq 0\end{matrix}\right.\)

\(\Rightarrow \sqrt{x+4}+\sqrt{x-1}>2\)

Do đó pt \(\sqrt{x+4}+\sqrt{x-1}=2\) vô nghiệm

23 tháng 6 2018

Giải:

a) \(\sqrt{\left(x-3\right)^2}=3-x\)

\(\Leftrightarrow\left|x-3\right|=3-x\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=3-x\\x-3=x-3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+x=3+3\\x-x=-3+3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=6\\0x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\0x=0\end{matrix}\right.\)

Vậy ...

b) \(\sqrt{25-20x+4x^2}+2x=5\)

\(\Leftrightarrow\sqrt{5^2-2.5.2x+\left(2x\right)^2}+2x=5\)

\(\Leftrightarrow\sqrt{\left(5-2x\right)^2}+2x=5\)

\(\Leftrightarrow\left|5-2x\right|+2x=5\)

\(\Leftrightarrow\left|5-2x\right|=5-2x\)

\(\Leftrightarrow\left[{}\begin{matrix}5-2x=5-2x\\5-2x=2x-5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-2x+2x=5-5\\-2x-2x=-5-5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}0x=0\\-4x=-10\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}0x=0\\x=\dfrac{5}{2}\end{matrix}\right.\)

Vậy ...

c) \(\sqrt{1-12x+36x^2}=5\)

\(\Leftrightarrow\sqrt{\left(1-6x\right)^2}=5\)

\(\Leftrightarrow\left|1-6x\right|=5\)

\(\Leftrightarrow\left[{}\begin{matrix}1-6x=5\\1-6x=-5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}6x=1-5\\6x=1-\left(-5\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}6x=-4\\6x=6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\x=1\end{matrix}\right.\)

Vậy ...

7 tháng 8 2018

( x2 - 4x + 2 )2 + ( x2 - 4x -4 ) = 0

( x- 2 )2 - ( x2 + 4x +4 ) = 0

( x2 - 2 )2 - ( x2 + 2 )2 = 0 

(x2 -2 - x2 -2 ).( x2 -2 + x2 +2 ) = 0

-4 . 2x2 =0

-8x2      = 0 

   x2       = 0 

 => x    = 0 

Vậy x=0 

7 tháng 8 2018

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

<=>   \(\left(x^2-4x+2\right)^2+\left(x^2-4x+2\right)-6=0\)

Đặt:  \(x^2-4x+2=t\)khi đó pt trở thành:

\(t^2+t-6=0\)

<=> \(\left(t-2\right)\left(t+3\right)=0\)

<=> \(\orbr{\begin{cases}t=2\\t=-3\end{cases}}\)

đến đây về pt bậc 2 bạn tự làm nhé

11 tháng 2 2020

\(4x^2-21x+23+2\sqrt{x+1}=0\) (x\(\ge-1\))

\(\Leftrightarrow\left(4x^2-20x+25\right)-\left(x+1+2\sqrt{x+1}+1\right)\)=0

\(\Leftrightarrow\left(2x-5\right)^2=\left(\sqrt{x+1}+1\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-5=\sqrt{x+1}+1\\2x-5=-\sqrt{x+1}-1\end{matrix}\right.\) ....

11 tháng 2 2020

Ta có: \(4x^2-21x+23+2\sqrt{x+1}=0\left(Đkxđ:x\ge-1\right)\)

\(\Leftrightarrow4x^2-21x+23=-2\sqrt{x+1}\)

\(\Leftrightarrow16x^4+441x^2+529-168x^3+184x^2-966x=4\left(x+1\right)\)

\(\Leftrightarrow16x^4-168x^3+625x^2-970x+525=0\)

\(\Leftrightarrow\left(16x^3-120x^2+265x-175\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-\frac{5}{4}\right)\left(16x^2-100x+140\right)=0\)

.............................................................

11 tháng 1 2017

Bài 1:

\(A=4x^2+4x-1\)

\(=4x^2+4x+1-2\)

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

Dấu "=" xảy ra khi \(x=-\frac{1}{2}\)

Bài 2:

Bình phương 2 vế 

\(\sqrt{\left(3x^2-4x+3\right)^2}=\left(1-2x\right)^2\)

\(\Leftrightarrow3x^2-4x+3=4x^2-4x+1\)

\(\Leftrightarrow2-x^2\Leftrightarrow x^2=2\Leftrightarrow x=-\sqrt{2}\) (tm)

\(x=-\sqrt{a}\Rightarrow-\sqrt{2}=-\sqrt{a}\Rightarrow a=2\)

11 tháng 1 2017

4x^2+4x-1

=4x^2+4x+1-2

=(2x+1)^2-2

=> (2x+1)^2\(\ge\)0 voi moi x

=> (2x+1)^2 \(\ge\)2

=> GTNN la 2

a: \(A=\dfrac{1}{x-1}\cdot5\sqrt{3}\cdot\left|x-1\right|\cdot\sqrt{x-1}\)

\(=\dfrac{5\sqrt{3}}{x-1}\cdot\left(x-1\right)\cdot\sqrt{x-1}=5\sqrt{3}\cdot\sqrt{x-1}\)

b: \(B=10\sqrt{x}-3\cdot\dfrac{10\sqrt{x}}{3}-\dfrac{4}{x}\cdot\dfrac{x\sqrt{x}}{2}\)

\(=10\sqrt{x}-10\sqrt{x}-\dfrac{4\sqrt{x}}{2}=-2\sqrt{x}\)

c: \(C=x-4+\left|x-4\right|\)

=x-4+x-4

=2x-8