1.Tìm x biết
a)(2x-6).(x+4)=0 b)(x+3).(x2+1)=0 c)(4x+12).(x-15)=0 d)-x(x+7)=0
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25 tháng 3 2020
Bài 1:
a) (3x - 2)(4x + 5) = 0
<=> 3x - 2 = 0 hoặc 4x + 5 = 0
<=> 3x = 2 hoặc 4x = -5
<=> x = 2/3 hoặc x = -5/4
b) (2,3x - 6,9)(0,1x + 2) = 0
<=> 2,3x - 6,9 = 0 hoặc 0,1x + 2 = 0
<=> 2,3x = 6,9 hoặc 0,1x = -2
<=> x = 3 hoặc x = -20
c) (4x + 2)(x^2 + 1) = 0
<=> 4x + 2 = 0 hoặc x^2 + 1 # 0
<=> 4x = -2
<=> x = -2/4 = -1/2
d) (2x + 7)(x - 5)(5x + 1) = 0
<=> 2x + 7 = 0 hoặc x - 5 = 0 hoặc 5x + 1 = 0
<=> 2x = -7 hoặc x = 5 hoặc 5x = -1
<=> x = -7/2 hoặc x = 5 hoặc x = -1/5
M
1 tháng 4 2020
e, 3x(2-x) =15(x-2)
\(\Leftrightarrow3x\left(2-x\right)-15\left(x-2\right)=0\)
\(\Leftrightarrow-3x\left(x-2\right)-15\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(-3x-15\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\-3x-15=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)
Vậy..
f, (x+5)(x+4)=0
\(\Leftrightarrow\left\{{}\begin{matrix}x+5=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-5\\x=-4\end{matrix}\right.\)
Vậy..
g, x(x+4)=0
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)
,h, (2x -4)(x-2)=0
\(\Leftrightarrow2\left(x-2\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2-1\right)=0\)
\(\Leftrightarrow x-2=0\Leftrightarrow x=2\)
i, (x+1/5)(2x-3)=0
\(\Leftrightarrow\left\{{}\begin{matrix}x+\frac{1}{5}=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{-1}{5}\\x=\frac{3}{2}\end{matrix}\right.\)
k, x²-4x=0
\(\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
m, 4x²-1=0
\(\Leftrightarrow\left(2x\right)^2-1^2=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-1=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=1\\2x=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{2}\\x=\frac{-1}{2}\end{matrix}\right.\)
n, x²-6x+9=0
\(\Leftrightarrow x^2-2.x.3+3^2=0\)
\(\Leftrightarrow\left(x-3\right)^2=0\Leftrightarrow x-3=0\)
<=> x=3
l, (3x-5)²-(x+4)²=0
\(\Leftrightarrow\left(3x-5-x-4\right)\left(3x-5+x+4\right)=0\)
\(\Leftrightarrow\left(2x-9\right)\left(4x-1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-9=0\\4x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=9\\4x=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{9}{2}\\x=\frac{1}{4}\end{matrix}\right.\)
Vậy ..
o, 7x(x+2)-5(x+2)=0
\(\Leftrightarrow\left(x+2\right)\left(7x-5\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+2=0\\7x-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\7x=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-2\\x=\frac{5}{7}\end{matrix}\right.\)
Vậy....
p, 3x(2x-5)-4x+10=0
\(\Leftrightarrow3x\left(2x-5\right)-\left(4x-10\right)=0\)
\(\Leftrightarrow3x\left(2x-5\right)-2\left(2x-5\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-5=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=5\\3x=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{5}{2}\\x=\frac{2}{3}\end{matrix}\right.\)
Vậy...
q, (2-2x)-x²+1=0
\(\Leftrightarrow2\left(1-x\right)-\left(x^2-1^2\right)=0\)
\(\Leftrightarrow2\left(1-x\right)-\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow2\left(1-x\right)+\left(1-x\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(1-x\right)\left(2+x+1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}1-x=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)
Vậy ....
r, x(1-3x)=5(1-3x)
\(\Leftrightarrow x\left(1-3x\right)-5\left(1-3x\right)=0\)
\(\Leftrightarrow\left(1-3x\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}1-3x=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-3x=-1\\x=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{3}\\x=5\end{matrix}\right.\)
s, 2x-3/4+x+1/6=3
\(\Leftrightarrow x-\frac{7}{12}=3\Leftrightarrow x=3+\frac{7}{12}=\frac{43}{12}\)
30 tháng 1 2017
a.
\(\left|x+10\right|=15\Rightarrow\orbr{\begin{cases}x+10=15\\x+10=-15\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=-25\end{cases}}}\)
b.
\(\left|x-3\right|+5=7\Rightarrow\left|x-3\right|=2\Rightarrow\orbr{\begin{cases}x-3=2\\x-3=-2\end{cases}}\Rightarrow\orbr{\begin{cases}x=5\\x=1\end{cases}}\)
c.
\(\left|x-3\right|+12=6\Rightarrow\left|x-3\right|=-6\Rightarrow x=\Phi\)
Phương trình vô nghiệm
d.
\(\left(2x+4\right)\left(3x-9\right)=0\Rightarrow\orbr{\begin{cases}2x+4=0\\3x-9=0\end{cases}\Rightarrow}\orbr{\begin{cases}2x=-4\\3x=9\end{cases}}\Rightarrow\orbr{\begin{cases}x=-2\\x=3\end{cases}}\)
e.
\(x^2-5x=0\Rightarrow x\left(x-5\right)=0\Rightarrow\orbr{\begin{cases}x=0\\x-5=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=5\end{cases}}\)
f.
\(\left(x+3\right)\left(4-2x\right)=70\Rightarrow4x-2x^2+7-6x=70\Rightarrow2x^2+2x+63=0\Rightarrow2\left(x+\frac{1}{2}\right)^2+\frac{123}{2}=0\)(vô lí)
Vậy phương trình vô nghiệm
1
(2x-6).(x+4)=0
=>2x-6=0 hoặc x+4=0
2x =0+6 x =0-4
2x =6 x =-4
x =6:2
x =3
Vậy x\(\in\){3;-4}
(x+3).(x2+1)=0
=>x+3=0 hoặc x2-1=0
x =0-3 x2 =0+1
x =-3 x2 =1
x2 =12
=>x=1
Vậy x\(\in\){-3;1}
(4x+12).(x-15)=0
=>4x+12=0 hoặc x-15=0
4x =0-12 x =0+15
4x =-12 x =15
x =-12:4
x =-3
Vậy x\(\in\){-3;15}
-x.(x+7)=0
=>-x=0 hoặc x+7=0
=>x=0 x =0-7
x =-7
Vậy x\(\in\){0;-7}
Chúc bn học tốt
\(a)\left(2x-6\right)\left(x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-6=0\\x+4=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=6\\x=0-4\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=-4\end{cases}}\)
\(b)\left(x+3\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+3=0\\x^2+1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-3\\x^2=-1\left(loai\right)\end{cases}}\)
\(c)\left(4x+12\right)\left(x-15\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}4x+12=0\\x-15=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}4x=-12\\x=0+15\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-3\\x=15\end{cases}}\)
\(d)-x\left(x+7\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}-x=0\\x+7=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-7\end{cases}}\)