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a. (3x - 1).(2x + 7) - (x + 1).(6x - 5) = 16
<=> 6x^2 + 19x - 7 - (6x^2 + x - 5) = 16
<=> 18x - 2 = 16
<=> 18x = 18
<=> x = 1
b. (10x + 9).x - (5x - 1).(2x + 3) = 8
<=> 10x^2 + 9x - (10x^2 + 13x - 3) = 8
<=> -4x + 3 = 8
<=> -4x = 5
<=> x = -5/4
c. (3x - 5).(7 - 5x) + (5x + 2).(3x - 2) - 2 = 0
<=> -15x^2 + 46x - 35 + 15x^2 - 4x - 4 - 2 = 0
<=> 42x - 41 = 0
<=> x = 41/42
![](https://rs.olm.vn/images/avt/0.png?1311)
a) (3x - 1)(2x + 7) - (x + 1)(6x - 5) = 16
6x2 + 21x - 2x - 7 - 6x2 + 5x - 6x + 5 = 16
(6x2 - 6x2) + (21x - 2x + 5x - 6x) + (-7 + 5) = 16
18x - 2 = 16
18x = 18
x = 1
Vậy x = 1
b) (10x + 9)x - (5x - 1)(2x + 3) = 8
10x2 + 9x - 10x2 - 15x + 2x + 3 = 8
(10x2 - 10x2) + (9x - 15x + 2x) + 3 = 8
-4x + 3 = 8
-4x = 5
x = \(\frac{-5}{4}\)
Vậy x = \(\frac{-5}{4}\)
c) x(x + 1)(x + 6) - x3 = 5x
(x2 + x)(x + 6) - x3 = 5x
x3 + 7x2 + 6x - x3 = 5x
7x2 + 6x = 5x
x(7x + 6) = 5x
=> 7x + 6 = 5
7x = -1
x = \(\frac{-1}{7}\)
Vậy x = \(\frac{-1}{7}\)
d) (3x - 5)(7 - 5x) + (5x + 2)(3x - 2) - 2 = 0
21x - 15x2 - 35 + 25x + 15x2 - 10x + 6x - 4 - 2 = 0
(-15x2 + 15x2) + (21x + 25x - 10x + 6x) + (-35 - 4 - 2) = 0
42x - 41 = 0
42x = 41
x = \(\frac{41}{42}\)
Vậy x = \(\frac{41}{42}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a/ pt đãcho tương đương với
6x\(^2\)+ 21x -2x-7-6x+5x-6x+5= 16
<=>18x=18
=> x=1
b/ pt đã cho tương đương với
10x\(^2\)+9x-10x\(^2\)-15x+2x+3= 8
<=> -4x=5
<=.> x=-\(\frac{5}{4}\)
c/ pt đã cho tương đương với
21x-15x\(^2\)-35+25x+15x\(^2\)-10x+6x-4-2=0
<=>42x=41
<=> x= \(\frac{41}{42}\)
d/ pt đã cho tương đương với
( x\(^2\)+x )(x+6)-x\(^3\)=5x
<=> x\(^3\)+6x\(^2\)+x\(^2\)+6x-x\(^3\)=5x
<=> 8x\(^2\)+6x-5x=0
<=>8x\(^2\)+16x-10x-5x=0
<=> (x+2)2x-5(x+2)=0
<=> (x+2)(2x-5)=0
<=>x+2=0 hoặc 2x+5=0
=> x=-2 hoặc x= -\(\frac{5}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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, \(\left(5x+3\right)\left(4x-7\right)-\left(10x+9\right)\left(2x-3\right)=10\Leftrightarrow-11x+6=10\Leftrightarrow x=-\frac{4}{11}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(4-3x\right)\left(10x-5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}4-3x=0\\10x-5=0\end{cases}\Rightarrow\orbr{\begin{cases}3x=4\\10x=5\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{4}{3}\\x=\frac{1}{2}\end{cases}}}\)
\(\left(7-2x\right)\left(4+8x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}7-2x=0\\4+8x=0\end{cases}\Rightarrow\orbr{\begin{cases}2x=7\\8x=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{1}{2}\end{cases}}}}\)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
một đòn bẫy dài một mét .đặt ở đâu để có thể dùng 3600n có thể nâng tảng đá nặng 120kg?
a) Ta có: \(3x\left(x+1\right)-2x\left(x+20\right)=-1-x\)
\(\Leftrightarrow3x^2+3x-2x^2-40x+1+x=0\)
\(\Leftrightarrow x^2-36x+1=0\)
\(\Leftrightarrow x^2-36x+324-323=0\)
\(\Leftrightarrow\left(x-18\right)^2=323\)
\(\Leftrightarrow\left[{}\begin{matrix}x-18=\sqrt{323}\\x-18=-\sqrt{323}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=18+\sqrt{323}\\x=18-\sqrt{323}\end{matrix}\right.\)
Vậy: \(x\in\left\{18+\sqrt{323};18-\sqrt{323}\right\}\)
b) Ta có: \(\left(3x-1\right)\left(2x+7\right)-\left(x+1\right)\left(6x-5\right)=16\)
\(\Leftrightarrow6x^2+21x-2x-7-\left(6x^2-5x+6x-5\right)-16=0\)
\(\Leftrightarrow6x^2+19x-7-\left(6x^2+x-5\right)-16=0\)
\(\Leftrightarrow6x^2+19x-7-6x^2-x+5-16=0\)
\(\Leftrightarrow18x-18=0\)
\(\Leftrightarrow18x=18\)
hay x=1
Vậy: x=1
c) Ta có: \(\left(10x+9\right)\cdot x-\left(5x-1\right)\left(2x+3\right)=8\)
\(\Leftrightarrow10x^2+9x-\left(10x^2+15x-2x-3\right)-8=0\)
\(\Leftrightarrow10x^2+9x-10x^2-13x+3-8=0\)
\(\Leftrightarrow-4x-5=0\)
\(\Leftrightarrow-4x=5\)
hay \(x=\frac{-5}{4}\)
Vậy: \(x=\frac{-5}{4}\)