Giải các phưong trình tích sau : a.x-5=1/3(x+2) b.x/3+x/4=1/5-x/6 c.x/2+x/3=1/4-x/5 d.(x+1)^2-5=x^2+11
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B1:
A.1+1=2
B.6+2+5=13
C.6+4+6+2+7=25
D.6+4+7+8=25
B2:
A.x+2=3
x=3-2
x=1
B.x+5=4+6
x+5=10
x=10-5
x=5
C.7=4+x
x=7-4
x=3
D.4+x={3+5+7}+{3+5}
4+x=15+8
4+x=23
x=23-4
x=19
Bài 1 :
A.1+1=2
B.6+2+5=13
C.6+4+6+2+7=25
D.6+4+7+8=25
Bài 2 :
A.x+2=3
x=3-2
x=1
B.x+5=4+6
x+5=10
x=10-5
x=5
C.7=4+x
x=7-4
x=3
D.4+x={3+5+7}+{3+5}
4+x=15+8
4+x=23
x=23-4
x=19
Chúc bạn học tốt nha ~_~
a) x2 - 5x - 6 = 0
=> x2 - 2x - 3x - 6 = 0
=> (x2 - 2x) + (-3x - 6) = 0
=> x(x - 2) - 3 (x - 2) = 0
=> (x - 2) (x - 3) = 0
=> x - 2 = 0 => x = 2
x - 3 = 0 => x = 3
còn lại tương tự nhé!! 46566578768698945635655675656788787868789789879789098089364556546
`a,(2x-5)(12+5x)=0`
\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\12+5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\5x=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{12}{5}\end{matrix}\right.\)
`b, (x-3)(x-4)-2(x-3)=0`
`<=>(x-3)(x-4-2)=0`
`<=>(x-3)(x-6)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=6\end{matrix}\right.\)
`c, x(x-1)(x+1)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
`d, (2x)/3 +(2x-1)/6=0`
`<=> (4x)/6 +(2x-1)/6=0`
`<=> (4x+2x-1)/6=0`
`<=> (6x-1)/6=0`
`<=> 6x-1=0`
`<=> 6x=1`
`<=>x=1/6` ( đề là vậy à bạn )
a) \(\left(2x-5\right)\left(12+5x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\12+5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\5x=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2,5\\x=-2,4\end{matrix}\right.\)
b) \(\left(x-3\right)\left(x-4\right)-2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left[\left(x-4\right)-2\right]=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-6\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=6\end{matrix}\right.\)
c) \(x\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-1=0\\x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=1\\x=0\end{matrix}\right.\)
d) \(\dfrac{2x}{3}+\dfrac{2x-1}{6}=0\)
\(\Leftrightarrow\dfrac{4x+2x-1}{6}=0\)
\(\Leftrightarrow6x-1=0\)
\(\Leftrightarrow6x=1\Leftrightarrow x=\dfrac{1}{6}\)
a>x+y=5=> y=5-x
\(!x+1!+!3-x!\ge!x+1+3-x!=4\)
đẳng thức khi -1<=x<=3
=> xem lại đề
a,x2.x3=25
=>x5=25
=>x=2
b,x+18=5.4^2
=>x=5.16-18
=>x=62
c,x.(x^2)^3=x^5
=>x.x5=x5
=>x=0,1
d,2x.7=224
2x=32
=>2x=25
=>x=5
e,(3x+5)2=289
=>(3x+5)2=172
=>3x+5=17
=>3x=12
=>x=4
g,32x+1.11=2673
=>32x=243
=>32x=35
=>x=\(\frac{5}{2}\)