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\(1.x^2+11x=0\)
\(\Leftrightarrow x\left(x+11\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+11=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-11\end{cases}}\)
\(2.\left(x^2-1\right)\left(x^2-9\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-1\right)\left(x+9\right)\left(x-9\right)=0\)
chia thành 4 TH :
\(TH1:X-1=0\)
\(\Leftrightarrow x=1\)
\(TH2:x+1=0\)
\(\Leftrightarrow x=-1\)
\(TH3:X+9=0\)
\(\Leftrightarrow X=-9\)
\(TH4:x-9=0\)
\(\Leftrightarrow x=9\)
Kết luận ....
\(3.\left(\left|x+1\right|-5\right)\left(x^2-9\right)\)
\(\Leftrightarrow\left(\left|x+1\right|-5\right)\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}\left|x+1\right|-5=0\\x-3=0\\x+3=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\left|x+1\right|=5\\x=3\\x=-3\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x+1=+_-5\Leftrightarrow x+1=5,x+1=-5\Leftrightarrow x=4,x=-6\\x=3x\\x=-3\end{cases}}\)
kết luận x=.....
\(4.\left(3x-16\right)⋮\left(x+2\right)\)
\(\Leftrightarrow\left(3x+6\right)-22\)
\(\Leftrightarrow3\left(x+2\right)-22⋮\left(x+2\right)\)
Vì\(\left(x+2\right)⋮\left(x+2\right)\)
\(\Rightarrow\left(3x-16\right)⋮\left(x+2\right)\)
Kết luận x=.....
\(\text{a/96-3(x+1)=42}\)
\(3\left(x+1\right)=54\)
\(x+1=54:3\)
\(x+1=18\)
\(\Rightarrow x=17\)
\(\text{b/2.x-18=20}\)
\(2x=38\)
\(x=38:2\)
\(x=19\)
\(\text{c/134-5.(x+4)=34}\)
\(5\left(x+4\right)=100\)
\(x+4=100:5\)
\(x+4=20\)
\(\Rightarrow x=16\)
học tốt
a) 96-3(x+1)=42
3(x+1)=96-42
3(x+1)=54
x+1=54:3
x+1=18
x=18-1
Vậy x=17
b) 2x-18=20
2x=20+18
2x=38
x=38:2
x=19
Vậy x=19.
c) 134-5(x+4)=34
5(x+4)=134-34
5(x+4)=100
x+4=100:5
x+4=20
x=20-4
x=16
Vậy x=16.
a) 96-3(x+1)=42
3(x+1)=96-42
3(x+1)=54
x+1=54:3
x+1=18
x=18-1
x=17
Vậy x=17
b)2.x-18=20
2.x =20+18
2.x =38
x =38:2
x =19
Vậy x=19
c)134-5.(x+4)=34
5.(x+4)=134-34
5.(x+4)=100
x+4 =100:5
x+4 =20
x =20-4
x =16
Vậy x=16
4.\(\left(2-x\right)\left(5-x\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}2-x=0\\5-x=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\\x=5\end{cases}}}\)
5.\(\left(4x-16\right)\left(8-2x\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}4x-16=0\\8-2x=0\end{cases}\Leftrightarrow\hept{\begin{cases}4x=16\\-2x=-8\end{cases}\Leftrightarrow}\hept{\begin{cases}x=4\\x=4\end{cases}}}\)
Câu 4:
( 2- x ) . ( 5 - x ) = 0
=> 2 - x = 0 hoặc 5 - x = 0
=> x = 2 hoặc x = 5
Vậy x \(\in\){ 2 ; 5 }
Cậu 5:
( 4x - 16 ) . ( 8 - 2x ) = 0
=> 4x - 16 = 0 hoặc 8 - 2x = 0
=> 4x = 16 hoặc 2x = 8
=> x = 4 hoặc x = 4
Vậy x = 4
\(\left(x-4\right)^9=49.\left(x-4\right)^7\\ =>\left(x-4\right)^9:\left(x-4\right)^7=49\\ =>\left(x-4\right)^2=49\\ =>\left[{}\begin{matrix}x-4=7\\x-4=-7\end{matrix}\right.\\ =>\left[{}\begin{matrix}x=11\\x=-3\end{matrix}\right.\)
(x - 4)9 = 49 . (x - 4)7
(x - 4)9 : (x - 4)7 = 49
(x - 4)2 = 72 = (-7)2
TH1 : TH2 :
(x - 4)2 = 72 (x - 4)2 = (-7)2
x - 4 = 7 x - 4 = -7
x = 7 + 4 x = -7 + 4
x = 11 x = -3
Vậy x = 11 Vậy x = -3
Lời giải:
a.
$x=\frac{-5}{6}-\frac{2}{3}=\frac{-3}{2}$
b.
$\frac{2}{3}x=\frac{1}{10}-\frac{1}{2}=\frac{-2}{5}$
$x=\frac{-2}{5}: \frac{2}{3}=\frac{-3}{5}$
c.
$\frac{7}{8}x=\frac{2}{9}-\frac{1}{3}=\frac{-1}{9}$
$x=\frac{-1}{9}: \frac{7}{8}=\frac{-8}{63}$
d.
$\frac{5}{7}: x=\frac{1}{6}-\frac{4}{5}=\frac{-19}{30}$
$x=\frac{5}{7}: \frac{-19}{30}=\frac{-150}{133}$
e.
$(\frac{2}{5}-1\frac{2}{3}):x=\frac{2}{5}+\frac{3}{5}=1$
$\frac{-19}{15}: x=1$
$x=\frac{-19}{15}:1 =\frac{-19}{15}$
f.
$(-\frac{3}{4}+x).2\frac{2}{3}=1$
$\frac{-3}{4}+x=1: 2\frac{2}{3}=\frac{3}{8}$
$x=\frac{3}{8}+\frac{3}{4}=\frac{9}{8}$
Ta có \(\dfrac{2}{1\cdot5}+\dfrac{2}{5\cdot9}+\dfrac{2}{9\cdot13}+...+\dfrac{2}{x\left(x+4\right)}=\dfrac{56}{113}\)
\(\dfrac{1}{2}\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+...+\dfrac{4}{x\left(x+4\right)}\right)=\dfrac{56}{113}\)
\(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{x}-\dfrac{1}{x+4}=\dfrac{56}{113}:\dfrac{1}{2}\)
\(1-\dfrac{1}{x+4}=\dfrac{112}{113}\)
\(\dfrac{1}{x+4}=1-\dfrac{112}{113}=\dfrac{1}{113}\)
x + 4 = 113 ⇒ x = 109
\(\dfrac{2}{1.5}+\dfrac{2}{5.9}+...+\dfrac{2}{x\left(x+4\right)}=\dfrac{56}{113}\)
Xét: \(A=\dfrac{2}{1.5}+\dfrac{2}{5.9}+...+\dfrac{2}{x\left(x+4\right)}\)
\(A=\dfrac{1}{2}.\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{x-4}-\dfrac{1}{x}+\dfrac{1}{x}-\dfrac{1}{x+4}\right)\)
\(A=\dfrac{1}{2}.\left(1-\dfrac{1}{x+4}\right)\)
Với \(A=\dfrac{56}{113}\) thì
\(\dfrac{1}{2}.\left(1-\dfrac{1}{x+4}\right)=\dfrac{56}{113}\)
\(\left(1-\dfrac{1}{x+4}\right)=\dfrac{112}{113}\)
\(\dfrac{1}{x+4}=\dfrac{1}{113}\)
\(x=109\)
a) \(\left(x^2-9\right)\cdot\left(4^x-16\right)=0\)
\(\Rightarrow x^2-9=0\)hoặc \(4^x-16=0\)
\(x^2=9\) \(4^x=16\)
\(x^2=\left(\pm3\right)^2\) \(4^x=4^2\)
\(\Rightarrow x=\pm3\)hoặc \(x=2\)
b) \(5^x+5^{x+2}=650\)
\(\Rightarrow5^x+5^x\cdot25=650\)
\(\Rightarrow5^x\cdot\left(1+25\right)=650\)
\(\Rightarrow5^x\cdot26=650\)
\(\Rightarrow5^x=650\div26=25\)
\(\Rightarrow5^x=5^2\)
\(\Rightarrow x=2\)Vậy \(x=2\)
c) \(2^{x+2}-2^x=96\)
\(2^x\cdot4-2^x=96\)
\(2^x\cdot\left(4-1\right)=96\)
\(2^x\cdot3=96\)
\(2^x=96\div3=32\)
\(2^x=2^5\)Vậy \(x=5\)