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\(a.\left(10x+9\right)x-\left(5x-1\right)\left(2x+3\right)=8.\)
\(\Leftrightarrow10x^2+9x-\left(10x^2+15x-2x-3\right)=8\)
\(\Leftrightarrow10x^2+9x-10x^2-13x+3=8\)
\(\Leftrightarrow-4x+3=8\)
\(\Leftrightarrow-4x=5\)
\(\Leftrightarrow x=-\frac{5}{4}\)
\(b.\left(3x-5\right)\left(7-5x\right)+\left(5x-2\right)\left(3x-2\right)-2=0\)
\(\Leftrightarrow21x-15x^2-35+25x+15x^2-10x+6x-4-2=0\)
\(\Leftrightarrow42x-41=0\)
\(\Leftrightarrow42x=41\)
\(\Leftrightarrow x=\frac{41}{42}\)
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= -\(\frac{5}{2}\)
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 = \(\frac{41}{42}\)
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
\(\left(4-3x\right)\left(10x-5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}4-3x=0\\10x-5=0\end{cases}\Rightarrow\orbr{\begin{cases}3x=4\\10x=5\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{4}{3}\\x=\frac{1}{2}\end{cases}}}\)
\(\left(7-2x\right)\left(4+8x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}7-2x=0\\4+8x=0\end{cases}\Rightarrow\orbr{\begin{cases}2x=7\\8x=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{1}{2}\end{cases}}}}\)
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) x2 + 10x - 2x - 20 = 0
=> x(x + 10) - 2(x + 10) = 0
=> (x - 2)(x + 10) = 0
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+10=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-10\end{cases}}\)
b) \(x^2-5x-24=0\)
\(\Rightarrow x^2-5x+\frac{25}{4}-\frac{121}{4}=0\)
\(\Rightarrow\left(x-\frac{5}{2}\right)^2=\frac{121}{4}\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x-\frac{5}{2}\right)^2=\left(-\frac{11}{2}\right)^2\\\left(x-\frac{5}{2}\right)^2=\left(\frac{11}{2}\right)^2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{5}{2}=\left(-\frac{11}{2}\right)\\x-\frac{5}{2}=\frac{11}{2}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{6}{2}=3\\x=\frac{16}{2}=8\end{cases}}\)
c) x2 - 8x + 3x - 24 = 0
=> x(x - 8) + 3(x - 8) = 0
=> (x + 3)(x - 8) = 0
\(\Leftrightarrow\orbr{\begin{cases}x+3=0\\x-8=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-3\\x=8\end{cases}}\)
Bài 3:
1. \(\left(x-1\right)\left(x+2\right)+5x-5=0\)
\(\Rightarrow\left(x-1\right)\left(x+2\right)+5\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(x+2+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x+7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\)
Vậy.......................
2. \(\left(3x+5\right)\left(x-3\right)-6x-10=0\)
\(\Rightarrow\left(3x+5\right)\left(x-3\right)-2\left(3x+5\right)=0\)
\(\Rightarrow\left(3x+5\right)\left(x-3-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}3x+5=0\\x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{3}\\x=5\end{matrix}\right.\)
Vậy........................
3. \(\left(x-2\right)\left(2x+3\right)-7x^2+14x=0\)
\(\Rightarrow\left(x-2\right)\left(2x+3\right)-7x\left(x-2\right)=0\)
\(\Rightarrow\left(x-2\right)\left(2x+3-7x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-2=0\\-5x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{3}{5}\end{matrix}\right.\)
Vậy............................
4, 5 tương tự nhé bn!
bài 3
1 (x-1)(x+2)+5x-5=0
=>(x-1)(x+2)+(5x-5)=o
=>(x-1)(x+2)+5(x-1)=0
=>(x-1)(x+2+5)=0
=>(x-1)(x+7)=0
=>\(\left[{}\begin{matrix}x-1=0\\x+7=0\end{matrix}\right.\) =>\(\left[{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\)
vậy x=1 hoặc x=-7
2. (3x+5)(x-3)-6x-10=0
=>(3x+5)(x-3)-(6x+10)=0
=>(3x+5)(x-3)-2(3x+5)=0
=>(3x+5)(x-3-2)=0
=>(3x+5)(x-5)=0
=>\(\left[{}\begin{matrix}3x+5=0\\x-5=0\end{matrix}\right.\)=>\(\left[{}\begin{matrix}x=-\dfrac{5}{3}\\x=5\end{matrix}\right.\)
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
a)5x ( 1 - 2x ) - 3x ( x + 18 ) = 0
5x - 10x^2 - 3x^2 - 54x =0
-13x^2 - 49x =0
(-13x - 49)x =0
\(\Rightarrow\orbr{\begin{cases}-13x-49=0\\x=0\end{cases}\Rightarrow\orbr{\begin{cases}-13x=49\\x=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{49}{13}\\x=0\end{cases}}}\)
Vậy x= -49/13 hoặc x=0
b)5x - 10x^2 - 3x^2 - 54 = 0
(câu b giống câu a)