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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 \(S=\left\{-\dfrac{5}{2}\right\}\)
a) ( 3x + 2 )( x - 1 ) - ( x + 2 )( 3x + 1 ) = 7
<=> 3x2 - x - 2 - ( 3x2 + 7x + 2 ) = 7
<=> 3x2 - x - 2 - 3x2 - 7x - 2 = 7
<=> -8x - 4 = 7
<=> -8x = 11
<=> x = -11/8
b) ( 6x + 5 )( 2x + 3 ) - ( 4x + 3 )( 3x - 2 ) = 8
<=> 12x2 + 28x + 15 - ( 12x2 + x - 6 ) = 8
<=> 12x2 + 28x + 15 - 12x2 - x + 6 = 8
<=> 27x + 21 = 8
<=> 27x = -13
<=> x = -13/27
c) 2x( x + 3 ) - ( x + 1 )( 2x + 1 ) - 5 = 9
<=> 2x2 + 6x - ( 2x2 + 3x + 1 ) - 5 = 9
<=> 2x2 + 6x - 2x2 - 3x - 1 - 5 = 9
<=> 3x - 6 = 9
<=> 3x = 15
<=> x = 5
d) ( 5x + 3 )( 4x - 7 ) - ( 10x + 9 )( 2x - 3 ) = 10
<=> 20x2 - 23x - 21 - ( 20x2 - 12x - 27 ) = 10
<=> 20x2 - 23x - 21 - 20x2 + 12x + 27 = 10
<=> -11x + 6 = 10
<=> -11x = 4
<=> x = -4/11
a, \(\left(3x+2\right)\left(x-1\right)-\left(x+2\right)\left(3x+1\right)=7\Leftrightarrow-8x-4=7\Leftrightarrow x=-\frac{11}{8}\)
b, \(\left(6x+5\right)\left(2x+3\right)-\left(4x+3\right)\left(3x-2\right)=8\Leftrightarrow27x+21=8\Leftrightarrow x=-\frac{13}{27}\)
c, \(2x\left(x+3\right)-\left(x+1\right)\left(2x+1\right)-5=9\Leftrightarrow3x-6=9\Leftrightarrow x=5\)
d, \(\left(5x+3\right)\left(4x-7\right)-\left(10x+9\right)\left(2x-3\right)=10\Leftrightarrow-11x+6=10\Leftrightarrow x=-\frac{4}{11}\)
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) \(\left(5x^2-2x+10\right)^2-\left(3x^2+10x-8\right)^2=0\)
\(\Leftrightarrow\left(5x^2-2x+10-3x^2-10x+8\right)\left(5x^2-2x+10+3x^2+10x-8\right)=0\)
\(\Leftrightarrow\left(2x^2-5x+18\right)\left(8x^2+8x+2\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{1}{2}\\x=3\end{matrix}\right.\)
\(a,\left(2x-1\right)^2=49\)
\(\left[{}\begin{matrix}2x-1=7\\2x-1=-7\end{matrix}\right.\)
\(\left[{}\begin{matrix}2x=8\\2x=-6\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)
\(b,\left(2x+7\right)^2=9\left(x+2\right)^2\)
\(4x^2+28x+49=9x^2+36x+36\)
\(4x^2+28x+49-9x^2-36x-36=0\)
\(-5x^2-8x+13=0\)
\(5x^2+13-5x-13=0\)
\(x\left(5x+13\right)-1\left(5x+13\right)=0\)
\(\left(x-1\right)\left(5x+13\right)=0\)
\(\left[{}\begin{matrix}x=1\\5x=-13\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=1\\x=-\frac{13}{5}\end{matrix}\right.\)
\(c,4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)
\(\left[2\left(2x+7\right)\right]^2-\left[3\left(x+3\right)\right]^2=0\)
\(\left(4x+14\right)^2-\left(3x+9\right)^2=0\)
\(4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)
\(x=-5\)
\(d,\left(5x-3\right)^2-\left(4x-7\right)^2=0\)
\(25x^2-30x+9-16x^2+56x-49=0\)
\(9x^2+26x-40=0\)
\(9x^2+36x-10x-40=0\)
\(9x\left(x+4\right)-10\left(x+4\right)=0\)
\(\left(9x-10\right)\left(x+4\right)=0\)
\(\left[{}\begin{matrix}9x-10=0\\x+4=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\frac{10}{9}\\x=-4\end{matrix}\right.\)
\(\left(4-3x\right)\left(10x-5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}4-3x=0\\10x-5=0\end{cases}\Rightarrow\orbr{\begin{cases}3x=4\\10x=5\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{4}{3}\\x=\frac{1}{2}\end{cases}}}\)
\(\left(7-2x\right)\left(4+8x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}7-2x=0\\4+8x=0\end{cases}\Rightarrow\orbr{\begin{cases}2x=7\\8x=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{1}{2}\end{cases}}}}\)
rồi thực hiện đến hết ...
Brainchild bé ngây thơ qus e , ko thực hiện đến hết như thế đc đâu :>
\(\left(x-3\right)\left(2x-1\right)=\left(2x-1\right)\left(2x+3\right)\)
\(2x^2-7x+3=4x^2+4x-3\)
\(2x^2-7x+3-4x^2-4x+3=0\)
\(-2x^2-11x+6=0\)
\(2x^2+11x-6=0\)
\(2x^2+12x-x-6=0\)
\(2x\left(x+6\right)-\left(x+6\right)=0\)
\(\left(x+6\right)\left(2x-1\right)=0\)
\(x+6=0\Leftrightarrow x=-6\)
\(2x-1=0\Leftrightarrow2x=1\Leftrightarrow x=\frac{1}{2}\)
\(3x-2x^2=0\)
\(x\left(2x-3\right)=0\)
\(x=0\)
\(2x-3=0\Leftrightarrow2x=3\Leftrightarrow x=\frac{3}{2}\)
Tự lm tiếp nha
a: Ta có: \(4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)
\(\Leftrightarrow\left(4x+14-3x-9\right)\left(4x+14+3x+9\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(7x+23\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{23}{7}\end{matrix}\right.\)
c: Ta có: \(\left(x-3\right)^2-4=0\)
\(\Leftrightarrow\left(x-5\right)\cdot\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)
b.
PT $\Leftrightarrow (5x^2-2x+10)^2-(3x^2+10x-8)^2=0$
$\Leftrightarrow (5x^2-2x+10-3x^2-10x+8)(5x^2-2x+10+3x^2+10x-8)=0$
$\Leftrightarrow (2x^2-12x+18)(8x^2+8x+2)=0$
$\Leftrightarrow (x^2-6x+9)(4x^2+4x+1)=0$
$\Leftrightarrow (x-3)^2(2x+1)^2=0$
$\Leftrightarrow (x-3)(2x+1)=0$
$\Leftrightarrow x-3=0$ hoặc $2x+1=0$
$\Leftrightarrow x=3$ hoặc $x=-\frac{1}{2}$
d.
$x^2-2x=24$
$\Leftrightarrow x^2-2x-24=0$
$\Leftrightarrow (x+4)(x-6)=0$
$\Leftrightarrow x+4=0$ hoặc $x-6=0$
$\Leftrightarrow x=-4$ hoặc $x=6$
a)4(x+2)-7(2x-1)+9(3x-4)=30 b)2(5x-8)-3(4x-5)=4(3x-4)+11
<=>4x+8-14x+7+27x-36=30 <=>10x-16-12x+15=12x-16+11
<=>17x-21=30 <=> -14x=-4 <=>x=2/7
<=>17x=51
<=>x=3
1.
<=> \(\left[{}\begin{matrix}4-3x=0\\10-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=2\end{matrix}\right.\)
2.
<=>\(\left[{}\begin{matrix}7-2x=0\\4+8x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
3.
<=>\(\left[{}\begin{matrix}9-7x=0\\11-3x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{7}\\x=\dfrac{11}{3}\end{matrix}\right.\)
4.
<=>\(\left[{}\begin{matrix}7-14x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=2\end{matrix}\right.\)
5.
<=>\(\left[{}\begin{matrix}\dfrac{7}{8}-2x=0\\3x+\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{16}\\x=-\dfrac{1}{9}\end{matrix}\right.\)
6,7. ko đủ điều kiện tìm
a) \(\left(2x-1\right)^2=49\)
\(\Leftrightarrow\orbr{\begin{cases}2x-1=7\\2x-1=-7\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=8\\2x=-6\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=4\\x=-3\end{cases}}\)