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\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)
\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)
Bài 4:
a: Ta có: \(3\left(2x-3\right)-2\left(x-2\right)=-3\)
\(\Leftrightarrow6x-9-2x+4=-3\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
\(\Leftrightarrow3x=13\)
hay \(x=\dfrac{13}{3}\)
c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)
\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
\(\Leftrightarrow-8x=-8\)
hay x=1
![](https://rs.olm.vn/images/avt/0.png?1311)
b, - ĐKXĐ : \(\left\{{}\begin{matrix}x-2\ne0\\x+2\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ne2\\x\ne-2\end{matrix}\right.\)
Ta có : \(\frac{5x}{x^2-4}-\frac{4}{x+2}=\frac{5}{x-2}\)
=> \(\frac{5x}{x^2-4}-\frac{4\left(x-2\right)}{x^2-4}=\frac{5\left(x+2\right)}{x^2-4}\)
=> \(5x-4\left(x-2\right)=5\left(x+2\right)\)
=> \(5x-4x+8=5x+10\)
=> \(5x-4x-5x=10-8\)
=> \(-4x=2\)
=> \(x=-\frac{1}{2}\) ( TM )
Vậy phương trình trên có tập nghiệm là \(S=\left\{-\frac{1}{2}\right\}\)
c, Ta có : \(x^4-15x^2+56=0\)
=> \(\left(x^2\right)^2-\frac{2.x^2.15}{2}+\frac{225}{4}-\frac{1}{4}=0\)
=> \(\left(x^2-\frac{15}{2}\right)^2=\frac{1}{4}\)
=> \(\left[{}\begin{matrix}x^2-\frac{15}{2}=\sqrt{\frac{1}{4}}\\x^2-\frac{15}{2}=-\sqrt{\frac{1}{4}}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x^2=\sqrt{\frac{1}{4}}+\frac{15}{2}=8\\x^2=-\sqrt{\frac{1}{4}}+\frac{15}{2}=7\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\sqrt{8}\\x=-\sqrt{8}\\x=\sqrt{7}\\x=-\sqrt{7}\end{matrix}\right.\)
Vậy phương trình trên có tập nghiệm là \(S=\left\{\sqrt{8};-\sqrt{8};\sqrt{7};-\sqrt{7}\right\}\)
a)
\(\frac{x-5x-1}{6}=\frac{8-3x}{4}\)
\(\Leftrightarrow\frac{4x-20x-4}{24}=\frac{48-18x}{24}\)
\(\Leftrightarrow\frac{-16x-4}{24}=\frac{48-18x}{24}\)
\(\Leftrightarrow\frac{-16x-4-48+18x}{24}=0\)
\(\Leftrightarrow\frac{2x-52}{24}=0\)
\(\Rightarrow2x-52=0\)
\(x=\frac{52}{2}=26\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Tìm x, biết:
1) 2x ( x - 5) - x ( 2x - 4 ) = 15
<=> 2x2 - 10x - 2x2 + 4x - 15 = 0
<=> -6x - 15 = 0
<=> -6x = 15
<=> x = -15/6
2) ( x +1)( x + 2 ) - ( x + 4 ) ( x + 3 ) = 6
<=> x2 + 2x + x + 2 - x2 - 3x - 4x - 12 - 6 = 0
<=> -4x = -16
<=> x = 4
3) 4x2 - 4x + 5 - x ( 4x - 3) = 1 - 2x
<=> 4x2 - 4x + 5 - 4x2 + 3x - 1 + 2x = 0
<=> x + 4 = 0
<=> x = -4
4) ( x + 3 ) ( 2x + 1 ) - 2x2 = 4x - 5
<=> 2x2 + x + 6x + 3 - 2x2 - 4x + 5 = 0
<=> 3x + 8 = 0
<=> 3x = -8
<=> x = -8/3
5) -4 ( 2x - 8 ) + ( 2x - 1 )( 4x + 3 ) = 0
<=> - 8x + 32 + 8x2 + 6x - 4x - 3 = 0
.......
6) -3 . (x-2) + 4 . (2x-6) - 7 . (x-9)= 5 . (3-2)
<=> -3x + 6 + 8x - 24 - 7x + 63 - 5 = 0
<=> -2x + 40 = 0
<=> -2x = -40
<=> x = 20
Còn lại tương tự ....
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(4\left(18-5x\right)-12\left(3x-7\right)=15\left(2x-16\right)-6\left(x+14\right)\)
\(\Rightarrow72-20x-36x+84=30x-240-6x-84\)
\(\Rightarrow-20x-36x-30x+6x=-240-84-72-84\)
\(\Rightarrow-80x=-480\Rightarrow x=6\)
b, \(5\left(3x+5\right)-4\left(2x-3\right)=5x+3\left(2x+12\right)+1\)
\(\Rightarrow15x+25-8x+12=5x+6x+36+1\)
\(\Rightarrow15x-8x-5x-6x=36+1-25-12\)
\(\Rightarrow-4x=0\Rightarrow x=0\)
c, \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)
\(\Rightarrow10x-16-12x+15=12x-16+11\)
\(\Rightarrow10x-12x-12x=-16+11+16-15\)
\(\Rightarrow-14x=-4\Rightarrow x=\dfrac{2}{7}\)
d, \(5x-3\left\{4x-2\left[4x-3\left(5x-2\right)\right]\right\}=182\)
\(\Rightarrow5x-3\left[4x-2\left(4x-15x+6\right)\right]=182\)
\(\Rightarrow5x-3\left(4x-8x+30x-12\right)=182\)
\(\Rightarrow5x-12x+24x-90x+36=182\)
\(\Rightarrow-73x=182-36\)
\(\Rightarrow-73x=146\Rightarrow x=-2\)
Chúc bạn học tốt!!!
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\Rightarrow72-20x-36x+84=30x-240-6x-84\)
\(\Rightarrow80x=480\Rightarrow x=6\)
b) \(\Rightarrow15x+25-8x+12=5x+6x+36+1\)
\(\Rightarrow4x=0\Rightarrow x=0\)
c) \(\Rightarrow10x-16-12x+15=12x-16+11\)
\(\Rightarrow14x=4\Rightarrow x=\dfrac{2}{7}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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 \(S=\left\{-\dfrac{5}{2}\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)
\(\Leftrightarrow10x-16-12x+15=12x-16+11\)
\(\Leftrightarrow-2x-1=12x-5\Leftrightarrow-14x=-4\Leftrightarrow x=\frac{2}{7}\)
\(\left(2x+3\right)\left(x-4\right)+\left(x-5\right)\left(x-2\right)=\left(3x-4\right)\left(x-4\right)\)
\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x-5x+10=3x^2-12x-4x+16\)
\(\Leftrightarrow3x^2-12x-2=3x^2-16x+16\)
\(\Leftrightarrow4x=18\Leftrightarrow x=\frac{9}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
20) -5-(x + 3) = 2 - 5x ⇔ -5 - x - 3 = 2 -5x ⇔ 4x = 10 ⇔ x = \(\frac{5}{2}\)
Vậy...
2 ( 5x-8 ) - 3 ( 4x - 5 ) = 4 ( 3x - 4 ) +11
<=> 10x - 16 - 12x + 15 = 12x - 16 + 11
<=> 14x = 4
<=> x = \(\frac{4}{14}\)
<=> x = \(\frac{2}{7}\)
Mk xong đầu tiên đó k mk nha nha nha^^