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AT
12 tháng 10 2017
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!
12 tháng 10 2017
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
=>\(\left[{}\begin{matrix}3x+5=0\\x-5=0\end{matrix}\right.\)=>\(\left[{}\begin{matrix}x=-\dfrac{5}{3}\\x=5\end{matrix}\right.\)
14 tháng 10 2018
1) \(2\left(x+2\right)-\left(3x+1\right)\left(x+2\right)=0\)
\(\left(x+2\right)\left(2-3x-1\right)=0\)
\(\left(x+2\right)\left(1-3x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+2=0\\1-3x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-2\\x=\frac{1}{3}\end{cases}}}\)
2) \(3x\left(x-3\right)-\left(2x-6\right)=0\)
\(3x\left(x-3\right)-2\left(x-3\right)=0\)
\(\left(x-3\right)\left(3x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\3x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=\frac{2}{3}\end{cases}}}\)
3) \(\left(2x-1\right)^2=\left(3x-5\right)^2\)
\(\left(2x-1\right)^2-\left(3x-5\right)^2=0\)
\(\left(2x-1-3x+5\right)\left(2x-1+3x-5\right)=0\)
\(\left(4-x\right)\left(5x-6\right)=0\)
\(\Rightarrow\orbr{\begin{cases}4-x=0\\5x-6=0\end{cases}\Rightarrow\orbr{\begin{cases}x=4\\x=\frac{6}{5}\end{cases}}}\)
4) \(\left(4x+3\right)\left(x-1\right)=x^2-1\)
\(\left(4x+3\right)\left(x-1\right)=\left(x+1\right)\left(x-1\right)\)
\(\left(4x+3\right)\left(x-1\right)-\left(x+1\right)\left(x-1\right)=0\)
\(\left(x-1\right)\left(4x+3-x-1\right)=0\)
\(\left(x-1\right)\left(3x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\3x+2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=\frac{-2}{3}\end{cases}}}\)
5) \(6-4x-\left(2x-3\right)\left(x-3\right)=0\)
\(-2\left(2x-3\right)-\left(2x-3\right)\left(x-3\right)=0\)
\(\left(2x-3\right)\left(-2-x+3\right)=0\)
\(\left(2x-3\right)\left(1-x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-3=0\\1-x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=1\end{cases}}}\)
6) \(2x^2-5x-7=0\)
\(2x^2+2x-7x-7=0\)
\(2x\left(x+1\right)-7\left(x+1\right)=0\)
\(\left(x+1\right)\left(2x-7\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+1=0\\2x-7=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-1\\x=\frac{7}{2}\end{cases}}}\)
7) \(x^2-x-12=0\)
\(x^2+3x-4x-12=0\)
\(x\left(x+3\right)-4\left(x+3\right)\)
\(\left(x+3\right)\left(x-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+3=0\\x-4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-3\\x=4\end{cases}}}\)
8) \(3x^2+14x-5=0\)
\(3x^2+15x-x-5=0\)
\(3x\left(x+5\right)-\left(x+5\right)=0\)
\(\left(x+5\right)\left(3x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+5=0\\3x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-5\\x=\frac{1}{3}\end{cases}}}\)
30 tháng 7 2018
1) -3x2+5x=0
-x(3x-5)=0
suy ra hoặc x=0 hoặc 3x-5=0. giải ra ta có nghiệm phương trình là 0 và 3/5
2) x2+3x-2x-6=0
x(x+3)-2(x+3)=0
(x-2)(x+3)=0
suy ra hoặc x-2=0 hoặc x+3=0. giải ra ta có nghiệm là 2 và -3
3) x2+6x-x-6=0
x(x+6)-(x+6)=0
(x-1)(x+6)=0. vậy nghiệm là 1 và -6
4) x2+2x-3x-6=0
x(x+2)-3(x+2)=0
(x-3)(x+2)=0
vậy nghiệm là -2 và 3
5) x(x-6)-4(x-6)=0
(x-4)(x-6)=0. vậy nghiệm là 4 và 6
6)x(x-8)-3(x-8)=0
(x-3)(x-8)=0
suy ra nghiệm là 3 và 8
7) x2-5x-24=0
x2-8x+3x-24=0
x(x-8)+3(x-8)=0
(x+3)(x-8)=0
vậy nghiệm là -3 và 8
LP
22 tháng 3 2020
câu 1: -3x2 + 5x = 0
suy ra -x(3x-5)=0
sung ra x = 0 hoặc 3x-5=0 suy ra 3x = 5 suy ra x = 5/3
VM
21 tháng 2 2020
Mình giải kĩ lại câu cuối nha.
\(\left(3x+5\right).\left(x^2+x+1\right)=0\)
+ Vì \(x^2+x+1>0\) \(\forall x.\)
\(\Rightarrow x^2+x+1\ne0.\)
\(\Leftrightarrow3x+5=0\)
\(\Leftrightarrow3x=0-5\)
\(\Leftrightarrow3x=-5\)
\(\Leftrightarrow x=\left(-5\right):3\)
\(\Leftrightarrow x=-\frac{5}{3}\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{-\frac{5}{3}\right\}.\)
Chúc bạn học tốt!
S
5 tháng 8 2016
1,
<=> \(\left(x-1\right)\left(x-2\right)^2=0\)
=> x=1 hoặc x=2
2,
<=>\(\left(x+1\right)\left(2x^2-3x+6\right)\)=0
=> x=-1
18 tháng 7 2017
1.
<=> ( x -1 ) ( x - 2 ) 2 = 0
=> x = 1 hoặc x = 2
2.
<=> ( x + 1 ) ( 2x2 - 3x + 6 ) = 0
=> x = -1
19 tháng 3 2020
a) Ta có: \(\left(x-1\right)\left(3x-6\right)=0\)
\(\Leftrightarrow\left(x-1\right)\cdot3\cdot\left(x-2\right)=0\)
Vì 3≠0
nên \(\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
Vậy: x∈{1;2}
b) Ta có: \(\left(2x+5\right)\left(1-3x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+5=0\\1-3x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-5\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-5}{2}\\x=\frac{1}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{-5}{2};\frac{1}{3}\right\}\)
c) Ta có: \(\left(x+1\right)\left(2x-3\right)\left(3x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\2x-3=0\\3x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\2x=3\\3x=5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\frac{3}{2}\\x=\frac{5}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{-1;\frac{3}{2};\frac{5}{3}\right\}\)
d) Ta có: \(6\left(x-2\right)\left(x-4\right)\left(1-7x\right)=0\)
Vì 6≠0
nên \(\left[{}\begin{matrix}x-2=0\\x-4=0\\1-7x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=4\\7x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=4\\x=\frac{1}{7}\end{matrix}\right.\)
Vậy: \(x\in\left\{2;4;\frac{1}{7}\right\}\)
e) Ta có: \(\left(x+1\right)^2\cdot\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x+1\right)^2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-2\end{matrix}\right.\)
Vậy: x∈{-1;-2}
f) Ta có: \(\left(3x-2\right)^2\cdot\left(x+1\right)\cdot\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(3x-2\right)^2=0\\x+1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\x=-1\\x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=2\\x=-1\\x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{2}{3}\\x=-1\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{2}{3};-1;2\right\}\)
g) Ta có: \(\left(5-x\right)^2\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(5-x\right)^2=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5-x=0\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\frac{1}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{5;\frac{1}{3}\right\}\)
h) Ta có: \(\left(14-2x\right)^2\cdot\left(3-x\right)\cdot\left(2x-4\right)=0\)
\(\Leftrightarrow4\left(7-x\right)^2\cdot\left(3-x\right)\cdot2\cdot\left(x-2\right)=0\)
\(\Leftrightarrow8\cdot\left(7-x\right)^2\cdot\left(3-x\right)\cdot\left(x-2\right)=0\)
Vì 8≠0
nên \(\left[{}\begin{matrix}\left(7-x\right)^2=0\\3-x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}7-x=0\\x=3\\x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=3\\x=2\end{matrix}\right.\)
Vậy: x∈{7;3;2}
i) Ta có: \(\left(5x-6\right)^2\cdot\left(x+2\right)\cdot\left(x+10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(5x-6\right)^2=0\\x+2=0\\x+10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x-6=0\\x=-2\\x=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=6\\x=-2\\x=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{6}{5}\\x=-2\\x=-10\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{6}{5};-2;-10\right\}\)
j) Ta có: \(\left(3x-3\right)^3\cdot\left(x+4\right)=0\)
\(\Leftrightarrow27\cdot\left(x-1\right)^3\cdot\left(x+4\right)=0\)
Vì 27≠0
nên \(\left[{}\begin{matrix}\left(x-1\right)^3=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-4\end{matrix}\right.\)
Vậy: x∈{1;-4}
12 tháng 10 2020
a) 2x (x-5) -(x2-10x +25)=0
\(\Leftrightarrow\)2x(x-5)-(x-5)2=0
\(\Leftrightarrow\)(x-5)(2x-x+5)=0
\(\Leftrightarrow\)(x-5)(x+5)=0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x-5=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\)
b) x2 - 9 +3x(x+3) = 0
\(\Leftrightarrow\)(x2 - 9) +3x(x+3) =0
\(\Leftrightarrow\)(x-3)(x+3)+3x(x+3)=0
\(\Leftrightarrow\)(x+3)(x-3+3x)=0
\(\Leftrightarrow\)(x+3)(4x-3)=0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x+3=0\\4x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=-3\\4x=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\frac{3}{4}\end{matrix}\right.\)
c) x3 - 16x = 0
\(\Leftrightarrow\)x(x2-16)=0
\(\Leftrightarrow\)x(x-4)(x+4)=0
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
d) (2x+3)(x-2) - (x2 -4x+4) = 0
\(\Leftrightarrow\)(2x+3)(x-2) -(x-2)2=0
\(\Leftrightarrow\)(x-2)(2x+3-x+2)=0
\(\Leftrightarrow\)(x-2)(x+5)=0
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)
e) 9x2 -(x2 -2x +1)=0
\(\Leftrightarrow\)(3x)2-(x-1)2=0
\(\Leftrightarrow\)(3x-x+1)(3x+x-1)=0
\(\Leftrightarrow\)(2x+1)(4x-1)=0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}2x+1=0\\4x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}2x=-1\\4x=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{2}\\x=\frac{1}{4}\end{matrix}\right.\)
f)x3-4x2 -9x +36 = 0
\(\Leftrightarrow\)(x3-9x)-(4x2-36)=0
\(\Leftrightarrow\)x(x2-9)-4(x2-9)=0
\(\Leftrightarrow\)(x-4)(x2-9)=0
\(\Leftrightarrow\)(x-4)(x-3)(x+3)=0
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-3=0\\x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=4\\x=3\\x=-3\end{matrix}\right.\)
g) 3x - 6 = (x-1).(x-2)
\(\Leftrightarrow\)3(x-2)=(x-1)(x-2)
\(\Leftrightarrow\)x-1=3
\(\Leftrightarrow\)x=4
i) (x-2).(x+2) +(2x+1)2 =-5x.(x-3) =5 (?? đề sao vậy ??)
k) x2 -1 = (x-1).(2x+3)
\(\Leftrightarrow\)(x-1)(x+1)=(x-1)(2x+3)
\(\Leftrightarrow\)x+1=2x+3
\(\Leftrightarrow\)x-2x=3-1
\(\Leftrightarrow\)-x=2
\(\Leftrightarrow\)x=-2
l) (2x-1)2 +(x+3).(x-3) -5x(x-2)=6
\(\Leftrightarrow\)4x2-4x+1+x2-9-5x2+10x=6
\(\Leftrightarrow\)6x-8=6
\(\Leftrightarrow\)6x=14
\(\Leftrightarrow\)x=\(\frac{7}{3}\)
1. x(x + 1) - x2 + 1 = 0
<=> x(x + 1) - (x2 - 1) = 0
<=> x(x + 1) - (x + 1)(x - 1) = 0
<=> (x - x + 1)(x + 1) = 0
<=> x + 1 = 0\
<=> x = -1
2. 4x(x - 2) - 6 + 3x = 0
<=> 4x(x - 2) - (3x - 6) = 0
<=> 4x(x - 2) - 3(x - 2) = 0
<=> (4x - 3)(x - 2) = 0
<=> \(\left[{}\begin{matrix}4x-3=0\\x-2=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=2\end{matrix}\right.\)
3. x(x + 2) - 3(x + 2) = 0
<=> (x - 3)(x + 2) = 0
<=> \(\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)