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HH
7 tháng 8 2017
a)
<=> 10x - 35 + 16x - 10 = 5
<=> 10x + 16x = 5 + 35 + 10
<=> 26x = 50
<=> x = 50/26 = 25/13
13 tháng 7 2021
a, ( 2x - 3 )2- (2x + 1)2 = -3
4x2-12x+9-4x2+4x-1=-3
-8x-1=-3
-8x=-2
x=\(\frac{1}{4}\)
b, (5x - 1) 2 - (5x + 4)(5x - 4) = 7
25x2-10x+1-25x2+16=7
-10x+17=7
-10x=-10
x=1
c, ( x- 5)2 + (x-3)(x+3) - 2(x + 1)2=0
x2-10x+25+x2-9-2x2-4x-2=0
-14x+14=0
-14(x-1)=0
=>x-1=0
x=1
AD
13 tháng 7 2021
a) \(\left(2x-3\right)^2-\left(2x+1\right)^2=-3\)
\(\Leftrightarrow4x^2-12x+9-4x^2-4x-1=-3\)
\(\Leftrightarrow-16x+8=-3\)
\(\Leftrightarrow-16x=-11\)
\(\Leftrightarrow x=\frac{11}{16}\)
b)\(\left(5x-1\right)^2-\left(5x+4\right)\left(5x-4\right)=7\)
\(\Leftrightarrow25x^2-10x+1-25x^2+16=7\)
\(\Leftrightarrow-10x+17=7\)
\(\Leftrightarrow-10x=-10\)
\(\Leftrightarrow x=1\)
c)\(\left(x-5\right)^2+\left(x-3\right)\left(x+3\right)-2\left(x+1\right)^2=0\)
\(\Leftrightarrow x^2-10x+25+x^2-9-2\left(x^2+2x+1\right)=0\)
\(\Leftrightarrow2x^2-10x-16-2x^2-4x-2=0\)
\(\Leftrightarrow-14x-18=0\)
\(\Leftrightarrow-14x=18\)
\(\Leftrightarrow x=-\frac{9}{7}\)
#H
NH
10 tháng 4 2020
a, \(\left(x-2\right)^2-\left(x+3\right)^2-4\left(x+1\right)=5\)
\(\Leftrightarrow x^2-4x+4-\left(x^2+6x+9\right)-4x-4=5\)
\(\Leftrightarrow x^2-4x+4-x^2-6x-9-4x-4=5\)
\(\Leftrightarrow-14x-9=5\)
\(\Leftrightarrow-14x=14\)
\(\Leftrightarrow x=-1\)
Vậy....
b, \(\left(2x-3\right)\left(2x+3\right)-\left(x-1\right)^2-3x\left(x-5\right)=-44\)
\(\Leftrightarrow\left(2x\right)^2-3^2-\left(x^2-2x+1\right)-3x^2+15x=-44\)
\(\Leftrightarrow4x^2-9-x^2+2x-1-3x^2+15x=-44\)
\(\Leftrightarrow-10+17x=-44\)
\(\Leftrightarrow17x=-34\)
\(\Leftrightarrow x=-2\)
Vậy....
c, \(\left(5x+1\right)^2-\left(5x+3\right)\left(5x-3\right)=30\)
\(\Leftrightarrow\left(5x\right)^2+10x+1-\left[\left(5x\right)^2-3^2\right]=30\)
\(\Leftrightarrow\left(5x\right)^2+10x+1-\left(5x\right)^2+9=30\)
\(\Leftrightarrow10x+10=30\)
\(\Leftrightarrow10x=20\)
\(\Leftrightarrow x=2\)
Vậy....
d, \(\left(x+3\right)^2+\left(x-2\right)\left(x+2\right)-2\left(x-2\right)^2=7\)
\(\Leftrightarrow x^2+6x+9+x^2-4-2\left(x^2-4x+4\right)=7\)
\(\Leftrightarrow2x^2+6x+5-2x^2+8x-8=7\)
\(\Leftrightarrow14x-3=7\)
\(\Leftrightarrow14x=10\)
\(\Leftrightarrow x=\frac{10}{14}=\frac{5}{7}\)
Vậy...
NT
0
19 tháng 6 2021
a) đk: x khác 1; \(\dfrac{3}{2}\)
\(P=\left[\dfrac{2x}{\left(2x-3\right)\left(x-1\right)}-\dfrac{5}{2x-3}\right]:\left(\dfrac{3-3x+2}{1-x}\right)\)
= \(\dfrac{2x-5\left(x-1\right)}{\left(2x-3\right)\left(x-1\right)}:\dfrac{5-3x}{1-x}\)
= \(\dfrac{-3x+5}{\left(2x-3\right)\left(x-1\right)}.\dfrac{1-x}{-3x+5}=\dfrac{-1}{2x-3}\)
b) Có \(\left|3x-2\right|+1=5\)
<=> \(\left|3x-2\right|=4\)
<=> \(\left[{}\begin{matrix}3x-2=4< =>x=2\left(Tm\right)\\3x-2=-4< =>x=\dfrac{-2}{3}\left(Tm\right)\end{matrix}\right.\)
TH1: Thay x = 2 vào P, ta có:
P = \(\dfrac{-1}{2.2-3}=-1\)
TH2: Thay x = \(\dfrac{-2}{3}\)vào P, ta có:
P = \(\dfrac{-1}{2.\dfrac{-2}{3}-3}=\dfrac{3}{13}\)
c) Để P > 0
<=> \(\dfrac{-1}{2x-3}>0\)
<=> 2x - 3 <0
<=> x < \(\dfrac{3}{2}\) ( x khác 1)
d) P = \(\dfrac{1}{6-x^2}\)
<=> \(\dfrac{-1}{2x-3}=\dfrac{1}{6-x^2}\)
<=> \(\dfrac{-1}{2x-3}=\dfrac{-1}{x^2-6}\)
<=> 2x - 3 = x2 - 6
<=> x2 - 2x - 3 = 0
<=> (x-3)(x+1) = 0
<=> \(\left[{}\begin{matrix}x=-1\left(Tm\right)\\x=3\left(Tm\right)\end{matrix}\right.\)
DP
7 tháng 7 2017
\(2x^2+3\left(x-1\right)\left(x+1\right)=5x\left(x+1\right)\)
\(2x^2+3.\left(x^2-1\right)=5x^2+5x\)
\(2x^2+3x^2-3=5x^2+5x\)
\(5x^2-3=5x^2+5x\)
\(5x=-3\)
\(\Rightarrow x=-\frac{3}{5}\)
7 tháng 7 2017
Ta có :
\(2x^2+3\left(x-1\right)\left(x+1\right)=5x\left(x+1\right)\)
=> \(2x^2+3\left(x^2-1\right)=5x^2+5x\)
=> \(2x^2+3x^2-3=5x^2+5x\)
=> \(5x^2-3=5x^2+5x\)
=> \(-3=5x\)
=> \(x=-\frac{3}{5}\)
NN
6
H
1 tháng 1 2019
\(x^2-y^2-5x+5y\)
\(=\left(x^2-y^2\right)-\left(5x-5y\right)\)
\(=\left(x+y\right)\left(x-y\right)-5\left(x-y\right)\)
\(=\left(x+y-5\right)\left(x-y\right)\)
29 tháng 8 2017
2x^2 + 3.( x^2 - 1^2 ) = 5x^2 + 5x
2x^2 + 3x^2 -3 = 5x^2 + 5x
5x^2 - 5x^2 -3 = 5x
-3 = 5x
x= \(\frac{-3}{5}\)
Lời giải:
$5x(x+1)=2x+2$
$\Leftrightarrow 5x(x+1)=2(x+1)$
$\Leftrightarrow 5x(x+1)-2(x+1)=0$
$\Leftrightarrow (x+1)(5x-2)=0$
$\Leftrightarrow x+1=0$ hoặc $5x-2=0$
$\Leftrightarrow x=-1$ hoặc $x=\frac{2}{5}$
5x(x + 1) = 2x + 2
<=> 5x(x + 1) = 2(x + 1)
<=> (x + 1)(5x - 2) = 0
<=> \(\left[{}\begin{matrix}x+1=0\\5x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{2}{5}\end{matrix}\right.\)