a) (x + 2 )2 = 9( x2 - 4x + 4 )
b) 4( 2x + 7 )2 - 9 ( x + 3 )2 = 0
c) ( 5x2 - 2x + 10 )2 = ( 3x2 + 10x -8)2
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25 tháng 3 2020
Bài 1:
a) (3x - 2)(4x + 5) = 0
<=> 3x - 2 = 0 hoặc 4x + 5 = 0
<=> 3x = 2 hoặc 4x = -5
<=> x = 2/3 hoặc x = -5/4
b) (2,3x - 6,9)(0,1x + 2) = 0
<=> 2,3x - 6,9 = 0 hoặc 0,1x + 2 = 0
<=> 2,3x = 6,9 hoặc 0,1x = -2
<=> x = 3 hoặc x = -20
c) (4x + 2)(x^2 + 1) = 0
<=> 4x + 2 = 0 hoặc x^2 + 1 # 0
<=> 4x = -2
<=> x = -2/4 = -1/2
d) (2x + 7)(x - 5)(5x + 1) = 0
<=> 2x + 7 = 0 hoặc x - 5 = 0 hoặc 5x + 1 = 0
<=> 2x = -7 hoặc x = 5 hoặc 5x = -1
<=> x = -7/2 hoặc x = 5 hoặc x = -1/5
9 tháng 7 2018
A. \(4\left(x+2\right)-7\left(2x-1\right)+9\left(3x-4\right)=30\)
\(\Leftrightarrow4x+8-14x+7+27x-36=30\)
\(\Leftrightarrow4x-14x+27x=30-8-7+36\)
\(\Leftrightarrow17x=51\)
\(\Leftrightarrow x=3\) . Vậy \(S=\left\{3\right\}\)
B. \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)
\(\Leftrightarrow10x-16-12x+15=12x-16+11\)
\(\Leftrightarrow10x-12x-12x=16-15-16+11\)
\(\Leftrightarrow10x=-4\)
\(\Leftrightarrow x=-\dfrac{2}{5}\) . Vậy \(S=\left\{-\dfrac{2}{5}\right\}\)
Câu C) bạn xem lại đề nha mik tính ko đc
D. \(\left(5x-3\right)4x-2x\left(10x-3\right)=15\)
\(\Leftrightarrow20x^2-12x-20x^2+6x=15\)
\(\Leftrightarrow-6x=15\)
\(\Leftrightarrow x=-\dfrac{5}{2}\) . Vậy \(S=\left\{-\dfrac{5}{2}\right\}\)
16 tháng 9 2021
9: \(\left(-2x\right)\left(3x^2-2x+4\right)=-6x^3+4x^2-8x\)
10 tháng 7 2019
\(1,\)\(\left(2x+3\right)^2=4x^2+12x+9\)
\(2,\)\(\left(3x+2y\right)^2=9x^2+12xy+4x^2\)
\(3,\)\(\left(3a-1\right)^2=9x^2-6x+1\)
\(4,\)\(\left(a-2\right)^2=a^2-4a+4\)
\(5,\)\(\left(1-5a\right)^2=1-10a+25a^2\)
\(6,\)\(\left(x-4\right)^3=x^3-12a^2+48a-64.\)
\(7,\)\(\left(x^2-2y\right)^2=x^4-4x^2y-4y^2\)
\(8,\)\(\left(5x^2-2\right)\left(5x^2+2\right)=25x^4-4\)
\(9,\)\(\left(2a^2-7\right)\left(2a^2+7\right)=4a^4-49\)
\(10,\)\(\left(x-1\right)\left(x^2+x+1\right)=x^3-1\)
\(11,\)\(\left(x^3-2\right)\left(x^6+2x^3+4\right)=x^9-8\)
\(12,\)\(\left(3x+2\right)\left(9x^2-6x+4\right)=27x^3+8\)
\(13,\)\(\left(x^2+3\right)\left(x^4-3x^2+9\right)=x^6+27\)
NN
10 tháng 7 2019
1, ( 2x + 3 )2 = 4x2 + 12x + 9
2, ( 3x + 2y )2 = 9x2 +12xy + 4y2
3 ( 3a - 1 )2 = 9a2 - 6x + 1
4, ( a - 2 )2 = a2 - 4a + 4
5, ( 1 - 5a )2 = 1 - 10a + 25a2
6, ( x- 4 )3 = x3 - 12x2 + 48x - 64
7, ( x2 - 2y )2 = x4 - 4x2y + 4y2
8, ( 5X2 - 2 ).( 5X2 + 2 ) = 25X2 - 4
9, ( 2a2 - 7 ).( 2a2 + 7 ) = 4a4 - 49
10, ( x - 1 ).( x2 + x + 1 ) = x3 - 1
31 tháng 3 2020
\(a,\left(2x-1\right)^2=49\)
\(\left[{}\begin{matrix}2x-1=7\\2x-1=-7\end{matrix}\right.\)
\(\left[{}\begin{matrix}2x=8\\2x=-6\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)
\(b,\left(2x+7\right)^2=9\left(x+2\right)^2\)
\(4x^2+28x+49=9x^2+36x+36\)
\(4x^2+28x+49-9x^2-36x-36=0\)
\(-5x^2-8x+13=0\)
\(5x^2+13-5x-13=0\)
\(x\left(5x+13\right)-1\left(5x+13\right)=0\)
\(\left(x-1\right)\left(5x+13\right)=0\)
\(\left[{}\begin{matrix}x=1\\5x=-13\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=1\\x=-\frac{13}{5}\end{matrix}\right.\)
31 tháng 3 2020
\(c,4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)
\(\left[2\left(2x+7\right)\right]^2-\left[3\left(x+3\right)\right]^2=0\)
\(\left(4x+14\right)^2-\left(3x+9\right)^2=0\)
\(4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)
\(x=-5\)
\(d,\left(5x-3\right)^2-\left(4x-7\right)^2=0\)
\(25x^2-30x+9-16x^2+56x-49=0\)
\(9x^2+26x-40=0\)
\(9x^2+36x-10x-40=0\)
\(9x\left(x+4\right)-10\left(x+4\right)=0\)
\(\left(9x-10\right)\left(x+4\right)=0\)
\(\left[{}\begin{matrix}9x-10=0\\x+4=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\frac{10}{9}\\x=-4\end{matrix}\right.\)
27 tháng 10 2021
Bài 1:
\(a,=15x^4-12x^3+9x^2\\ b,=-15x^3y^2+25x^2y^2-5xy^3\\ c,=5x^3-15x^2-4x^2+12x=5x^3-19x^2+12x\\ d,=3x^3-9x^2y+xy^2-3y^3+5x^2y-15xy^2=3x^3-3y^3-4x^2y-14xy^2\)
Bài 2:
\(a,=x^2+4x-21-x^2-4x+5=-16\\ b,=x^2+16x+64-2x^2-12x+32+x^2-4x+4=100\\ c,=x^4-16x^2-x^4+1=1-16x^2\\ d,=x^3+1-x^3+1=2\)
a) \(\left(x+2\right)^2-9\left(x-2\right)^2=0\)
\(\Leftrightarrow\left(x+2\right)^2-\left(3x-6\right)^2=0\)
\(\Leftrightarrow\left(x+2+3x-6\right)\left(x+2-3x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(4x-4\right)=0\\\left(8-2x\right)=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\end{matrix}\right.\)
b)\(4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)
\(\Leftrightarrow\left(4x+14\right)^2-\left(3x+9\right)^2=0\)
\(\Leftrightarrow\left(4x+14-3x-9\right)\left(4x+14+3x+9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\frac{23}{7}\end{matrix}\right.\)
c) \(\left(5x^2-2x+10\right)^2-\left(3x^2+10x-8\right)^2=0\)
\(\Leftrightarrow\left(5x^2-2x+10-3x^2-10x+8\right)\left(5x^2-2x+10+3x^2+10x-8\right)=0\)
\(\Leftrightarrow\left(2x^2-5x+18\right)\left(8x^2+8x+2\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{1}{2}\\x=3\end{matrix}\right.\)
Câu c sai một tí, chắc nhầm =))