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5 tháng 11 2017
Giải như sau.
(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y
⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn !
30 tháng 9 2018
\(\left(x+6\right)\left(2x+1\right)=0\)
<=> \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)
Vậy....
hk tốt
^^
DT
4
KN
5 tháng 9 2020
a. \(x\left(x-2\right)-x\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow x^2-2x-x^3+4x^2-3x=0\)
\(\Leftrightarrow-x^3+5x^2-5x=0\)
\(\Leftrightarrow-x\left(x^2-5x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}-x=0\\x^2-5x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\\left(x-\frac{5}{2}\right)^2-\frac{5}{4}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\\left(x-\frac{5}{2}\right)^2=\frac{5}{4}\end{cases}\Leftrightarrow}\hept{\begin{cases}x=0\\x-\frac{5}{2}=\frac{\sqrt{5}}{2}\\x-\frac{5}{2}=-\frac{\sqrt{5}}{2}\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=0\\x=\frac{5+\sqrt{5}}{2}\\x=\frac{5-\sqrt{5}}{2}\end{cases}}\)
HT
5 tháng 9 2020
a) \(x\left(x-2\right)-x\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow x\left(x-2-x^2+4x-3\right)=0\)
\(\Leftrightarrow x\left(-x^2+5x-5\right)=0\)
\(\Leftrightarrow x\left(x-\frac{5+\sqrt{5}}{2}\right)\left(x-\frac{5-\sqrt{5}}{2}\right)=0\)
=> \(x\in\left\{0;\frac{5+\sqrt{5}}{2};\frac{5-\sqrt{5}}{2}\right\}\)
b) \(\left(2x-5\right)\left(x+3\right)-\left(x-1\right)\left(2x+3\right)=0\)
\(\Leftrightarrow2x^2+x-15-2x^2-x+3=0\)
\(\Leftrightarrow-12=0\left(vn\right)\)
c) \(\left(x-2\right)\left(x^2+2x+8\right)-x^3-2x+1=0\)
\(\Leftrightarrow x^3+4x-16-x^3-2x+1=0\)
\(\Leftrightarrow2x=15\)
\(\Rightarrow x=\frac{15}{2}\)
LT
5
NC
19 tháng 2 2021
\(a,\left(2x-1\right)^2-\left(2x+3\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x-1-2x-3\right)=0\)
\(\Leftrightarrow-4\left(2x-1\right)=0\)
\(\Leftrightarrow2x-1=0\)
\(\Leftrightarrow2x=1\)
\(\Leftrightarrow x=\frac{1}{2}\)
\(b,\left(x+5\right)\left(x-2\right)-\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left(x^2+3x-10\right)-\left(x^2-9\right)=0\)
\(\Leftrightarrow x^2+3x-10-x^2+9=0\)
\(\Leftrightarrow3x-1=0\)
\(\Leftrightarrow3x=1\)
\(\Leftrightarrow x=\frac{1}{3}\)
XO
19 tháng 2 2021
a) (2x - 1)2 - (2x + 3)(2x - 1) = 0
<=> (2x - 1)(2x - 1 - 2x - 3) = 0
<=> (2x - 1).(-4) = 0
<=> 2x - 1 = 0
<=> x = 1/2
Vậy x = 1/2 là nghiệm phương trình
b) Ta có (x - 5)(x - 2) - (x - 3)(x + 3) = 0
<=> x2 - 7x + 10 - x2 + 9 = 0
<=> -7x + 19 = 0
<=> -7x = - 19
<=> x = 19/7
Vây x = 19/7 là nghiệm phương trình
5 tháng 9 2019
a) 3x(4x - 3) - 2x(5 - 6x) = 0
=> 6x2 - 9x - 10x + 12x2 = 0
=> 18x2 - 19x = 0
=> x(18x - 19) = 0
=> \(\orbr{\begin{cases}x=0\\18x-19=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=\frac{19}{18}\end{cases}}\)
b) 5(2x - 3) + 4x(x - 2) + 2x(3 - 2x) = 0
=> 10x - 15 + 4x2 - 8x + 6x - 4x2 = 0
=> 8x - 15 = 0
=> 8x = 15
=> x = 15 : 8 = 15/8
c) 3x(2 - x) + 2x(x - 1) = 5x(x + 3)
=> 6x - 3x2 + 2x2 - 2x = 5x2 + 15x
=> 4x - x2 - 5x2 - 15x = 0
=> -6x2 - 11x = 0
=> -x(6x - 11) = 0
=> \(\orbr{\begin{cases}-x=0\\6x-11=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=\frac{11}{6}\end{cases}}\)
2T
5 tháng 9 2019
a) \(3x\left(4x-3\right)-2x\left(5-6x\right)=0\)
\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)
\(\Leftrightarrow-19x=0\Leftrightarrow x=0\)
b) \(5\left(2x-3\right)+4x\left(x-2\right)+2x\left(3-2x\right)=0\)
\(\Leftrightarrow10x-15+4x^2-8x+6x-4x^2=0\)
\(\Leftrightarrow8x-15=0\Leftrightarrow x=\frac{15}{8}\)
30 tháng 3 2020
Bài 1:
a) Ta có: \(\left(1-2x\right)\left(1+2x\right)+\left(2x+3\right)^2=34\)
\(\Leftrightarrow1-4x^2+4x^2+12x+9-34=0\)
\(\Leftrightarrow12x-24=0\)
\(\Leftrightarrow12\left(x-2\right)=0\)
Vì 12≠0
nên x-2=0
hay x=2
Vậy: x=2
b) Ta có: \(\left(2x-3\right)^2+\left(3-2x\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(2x-3\right)^2-\left(2x-3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left[\left(2x-3\right)-\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(2x-3\right)\left(2x-3-x+1\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{3}{2}\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{3}{2};2\right\}\)
Bài 2:
a) Ta có: \(\frac{2x-5y}{x-y}-\frac{3y}{y-x}\)
\(=\frac{2x-5y}{x-y}+\frac{3y}{x-y}\)
\(=\frac{2x-5y+3y}{x-y}=\frac{2x-2y}{x-y}=\frac{2\left(x-y\right)}{x-y}=2\)
b) Ta có: \(\frac{x^2+3xy}{x^2-9y^2}+\frac{5x-x^2}{x^2-3xy}\)
\(=\frac{x\left(x+3y\right)}{\left(x-3y\right)\left(x+3y\right)}+\frac{x\left(5-x\right)}{x\left(x-3y\right)}\)
\(=\frac{x}{x-3y}+\frac{5-x}{x-3y}\)
\(=\frac{x+5-x}{x-3y}=\frac{5}{x-3y}\)
21 tháng 8 2021
a) \(\left(x+3\right)^2-\left(x-2\right)^3=\left(x+5\right)\left(x^2-5x+25\right)-108\)
\(\Leftrightarrow x^2+6x+9-x^2+4x-4=x^3-5x^2+25x+5x^2-25x+125-108\)
\(\Leftrightarrow x^3-10x+12=0\Leftrightarrow\left(x-2\right)\left(x^2+2x+6\right)=0\)
\(\Leftrightarrow x=2\)( do \(x^2+2x+6=\left(x+1\right)^2+4\ge4>0\))
a)
\(\left(1-2x\right)\cdot\left(1+2x\right)+\left(2x+3\right)^2=34\\ \Leftrightarrow1-4x^2+4x^2+12x+9=34\\ \Leftrightarrow12x+10=34\\ \Rightarrow x=\frac{34-10}{12}=2\)
b)
\(\left(2x-3\right)^2+\left(3-2x\right)\cdot\left(x-1\right)=0\\ \Leftrightarrow4x^2-12x+9+3x-3-2x^2+2x=0\\ \Leftrightarrow2x^2-7x+6=0\\ \Leftrightarrow6-7x+2x^2=0\\ \Leftrightarrow6-3x-4x+2x^2=0\\ \Leftrightarrow3\cdot\left(2-x\right)-2x\cdot\left(2-x\right)=0\\ \Leftrightarrow\left(3-2x\right)\cdot\left(2-x\right)=0\\\Rightarrow\left[{}\begin{matrix}3-2x=0\\2-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\frac{3}{2}\\x=2\end{matrix}\right.\)
\(a,\left(1-2x\right)\left(1+2x\right)+\left(2x+3\right)^2=34\)
\(1-4x^2+4x^2+12x+9=34\)
\(10+12x=34\)
\(12x=24\)
\(x=2\)
\(b,\left(2x-3\right)^2+\left(3-2x\right)\left(x-1\right)=0\)
\(4x^2+12x+9+3x-3-2x^2+2x=0\)
\(2x^2-7x+6=0\)
\(2x^2-3x-4x+6=0\)
\(x\left(2x-3\right)-2\left(2x-3\right)=0\)
\(\left(x-2\right)\left(2x-3\right)=0\)
\(\left[{}\begin{matrix}x-2=0\\2x-3=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=2\\x=\frac{3}{2}\end{matrix}\right.\)