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25 tháng 8 2016
a) ( 2x + 3 )^2 - 4( x - 1 )( x + 1 ) = 49
=>4x2+12x+9-4x2+4=49
=>12x+13=49
=>12x=36
=>x=3
b) 16x^2 - ( 4x - 5 )^2 = 15
=>16x2-16x2+40x-25=15
=>40x-25=15
=>40x=40
=>x=1
c) ( 2x + 1 )^2 - ( x - 1)^2 = 0
=>4x2+4x+1-x2+2x-1=0
=>3x2+6x=0
=>3x(x+2)=0
=>3x=0 hoặc x+2=0
=>x=0 hoặc x=-2
NT
26 tháng 8 2016
a) \(\left(2x+3\right)^2-4\left(x-1\right)\left(x+1\right)=49\\ =>4x^2+12x+9-4x^2+4=49\\=>12x+13=49\\ =>12x=36\\ =>x=3\)
b) \(16x^2-\left(4x-5\right)^2=15\\ =>16x^2-16x^2+40x-25=15\\ =>40x-25=15\\ =>40x=40\\ =>x=1\)
c) \(\left(2x+1\right)^2-\left(x-1\right)^2=0\\ =>4x^2+4x+1-x^2+2x-1=0\\ =>3x^2+6x=0\\ =>3x\left(x+2\right)=0\\ =>\left[\frac{3x=0}{x+2=0}\right]=>\left[\frac{x=0}{x=-2}\right]\)
TT
2
LA
9 tháng 7 2016
a) \(\left(2x-5\right)^2=49\)
\(\left(2x-5\right)^2=\left(\pm7\right)^2\)
\(=>2x-5=7\) hoặc \(2x-5=-7\)
\(\cdot2x-5=7\) \(\cdot2x-5=-7\)
\(2x=5+7\) \(2x=-7+5\)
\(2x=12\) \(2x=-2\)
\(x=12:2\) \(x=-2:2\)
\(x=6\) \(x=-1\)
Vậy x=6 hoặc x=-1
b/ \(\left(2x+5\right)^2-\left(1-2x\right)^2=10\)
\(4x^2+20x+25-\left(1-4x+4x^2\right)=10\)
\(4x^2+20x+25-1+4x-4x^2=10\)
\(24x+24=10\)
\(24x=10-24\)
\(24x=-14\)
\(x=\frac{-14}{24}\)
\(x=\frac{-7}{12}\)
c/ \(\left(9-2x\right)^3=27\)
\(\left(9-2x\right)^3=3^3\)
\(9-2x=3\)
\(2x=9-3\)
\(2x=6\)
\(x=6:2\)
\(x=3\)
14 tháng 8 2020
a) 16x^2 - (4x - 5)^2 = 15
<=> 16x^2 - 16x^2 + 40x - 25 = 15
<=> 40x = 40
<=> x = 1
b) (2x + 3)^2 - 4(x - 1)(x + 1) = 49
<=> 4x^2 + 12x + 9 - 4x^2 - 4x + 4x + 4 = 49
<=> 12x + 13 = 49
<=> 12x = 36
<=> x = 3
c) (2x + 1)(1 - 2x) + (1 - 2x)^2 = 18
<=> 1 - 4x^2 + 1 - 4x + 4x^2 = 18
<=> 2 - 4x = 18
<=> -4x = 16
<=> x = -4
d)2(x + 1)^2 - (x - 3)(x + 3) - (x - 4)^2 = 0
<=> 2x^2 + 4x + 2 - x^2 + 3^2 - x^2 + 8x - 16 = 0
<=> 12x - 5 = 0
<=> 12x = 5
<=> x = 5/12
e) (x - 5)^2 - x(x - 4) = 9
<=> x^2 - 10x + 25 - x^2 + 4x = 9
<=> -6x + 25 = 9
<=> -6x = 9 - 25
<=> -6x = -16
<=> x = -16/-6 = 8/3
f) (x - 5)^2 + (x - 4)(1 - x) = 0
<=> x^2 - 10x + 25 + x - x^2 - x - 4 + 4x = 0
<=> -5x + 21 = 0
<=> -5x = -21
<=> x = 21/5
26 tháng 9 2016
a ) \(9x^2-49=9\)
\(\Leftrightarrow9x^2=58\)
\(\Leftrightarrow x^2=29\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=29\\x=-29\end{array}\right.\)
Vậy ......................
b ) \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-1\right)\left(x+1\right)-27=0\)
\(\Leftrightarrow\left(x^3+3^3\right)-x.\left(x^2-1^2\right)-27=0\)
\(\Leftrightarrow x^3+27-x^3+x-27=0\)
\(\Leftrightarrow x=0\)
c ) \(\left(x-1\right)\left(x+2\right)-x-2=0\)
\(\Leftrightarrow x^2+2x-x-2-x-2=0\)
\(\Leftrightarrow x^2-4=0\)
\(\Leftrightarrow x^2=4\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=-2\end{array}\right.\)
Vây .....................
22 tháng 1 2019
a. \(\left(2x-1\right)\left(3x+2\right)\left(5-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\3x+2=0\\5-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{-2}{3}\\x=5\end{matrix}\right.\)
\(\Rightarrow S=\left\{\dfrac{1}{2};\dfrac{-2}{3};5\right\}\)
b. \(\left(2x+5\right)\left(x-4\right)=\left(x-5\right)\left(4-x\right)\)
\(\Leftrightarrow\left(2x+5\right)\left(x-4\right)+\left(x-5\right)\left(x-4\right)\)
\(\Leftrightarrow3x\left(x-4\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
\(\Rightarrow S=\left\{0;4\right\}\)
c. \(16x^2-8x+1=4\left(x+3\right)\left(4x-1\right)\)
\(\Leftrightarrow\left(4x-1\right)^2-4\left(x+3\right)\left(4x-1\right)=0\)
\(\Leftrightarrow\left(4x-1\right)\left(4x-1-4x-3\right)=0\)
\(\Leftrightarrow-4\left(4x-1\right)=0\Leftrightarrow4x-1=0\Leftrightarrow x=\dfrac{1}{4}\)
d. \(27x^2\left(x+3\right)-12\left(x^2+3x\right)=0\)
\(\Leftrightarrow27x^2\left(x+3\right)-12x\left(x+3\right)=0\)
\(\Leftrightarrow x\left(27x-12\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\27x-12=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4}{9}\\x=-3\end{matrix}\right.\)
\(\Rightarrow S=\left\{0;\dfrac{4}{9};-3\right\}\)
e. \(2\left(9x^2+6x+1\right)=\left(3x+1\right)\left(x-2\right)\)
\(\Leftrightarrow2\left(3x+1\right)^2-\left(3x+1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(6x+1-x+2\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(7x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\7x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{3}\\x=\dfrac{-3}{7}\end{matrix}\right.\)
\(\Rightarrow S=\left\{\dfrac{-1}{3};\dfrac{-3}{7}\right\}\)
g. \(\left(2x-1\right)^2=49\)
\(\Leftrightarrow2x-1=7\Leftrightarrow x=4\)
12 tháng 6 2017
a/ \(x^2+y^2=x^2+y^2+2xy-2xy =\left(x+y\right)^2-2xy\)
b/ mình không chắc nữa
bài 3
a/ \(9x^2-49=0 \Leftrightarrow x^2=\frac{49}{9} \Leftrightarrow\orbr{\begin{cases}x=\frac{7}{3}\\x=-\frac{7}{3}\end{cases}}\)
b/ \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x+1\right)\left(x-1\right)-27=0 \Leftrightarrow x^3+27-x\left(x^2-1\right)-27=0\)
\(\Leftrightarrow x^3-x^3+x=0\Leftrightarrow x=0\)
c/\(\left(x-1\right)\left(x+2\right)-x-2=0 \Leftrightarrow \left(x-1\right)\left(x+2\right)-\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)^2=0\Leftrightarrow\orbr{\begin{cases}x+2=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=1\end{cases}}}\)
d/ \(x\left(3x+2\right)+\left(x+1\right)^2-\left(2x-5\right)\left(2x+5\right)=0\)
\(\Leftrightarrow3x^2+2x+x^2+2x+1-4x^2+25=0\)
\(\Leftrightarrow4x+25=0 \Leftrightarrow x=\frac{-25}{4}\)
e/ mình lười qá ko viết đề đâu
\(\Leftrightarrow4x^2-7x-2-4x^2+4x+3=7\)
\(\Leftrightarrow-3x+1=7 \Leftrightarrow x=-2\)
có gì sai bn sửa lại nha
20 tháng 7 2022
b: \(\Leftrightarrow\dfrac{7x+10}{x+1}\left(x^2-x-2-2x^2+3x+5\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\left(-x^2+2x+3\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\left(x^2-2x-3\right)=0\)
=>(7x+10)(x-3)=0
hay \(x\in\left\{-\dfrac{10}{7};3\right\}\)
d: \(\Leftrightarrow\dfrac{13}{2x^2+7x-6x-21}+\dfrac{1}{2x+7}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\dfrac{13}{\left(2x+7\right)\left(x-3\right)}+\dfrac{1}{\left(2x+7\right)}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow26x+91+x^2-9-12x-14=0\)
\(\Leftrightarrow x^2+14x+68=0\)
hay \(x\in\varnothing\)
\(\left(2x+5\right)^2-2\left(2x+5\right)\left(x-1\right)+\left(x+1\right)^2=49\)
<=>\(\left\{\left(2x-5\right)-\left(x-1\right)\right\}^2=49\)
<=> \(\left(2x-5-x+1\right)^2=49\)
<=> \(\left(x-4\right)^2=49\)
<=> \(\hept{\begin{cases}x-4=7\\x-4=-7\end{cases}}\)
<=> \(\hept{\begin{cases}x=11\\x=-3\end{cases}}\)
học tốt