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x.y-x.2=0
=> x.y = 0 và x.2 = 0
=> x = 0 hoặc y = 0 và x = 0.
Vậy x = 0, y = 0
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1.x^2+11x=0\)
\(\Leftrightarrow x\left(x+11\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+11=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-11\end{cases}}\)
\(2.\left(x^2-1\right)\left(x^2-9\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-1\right)\left(x+9\right)\left(x-9\right)=0\)
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=.....
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(4x-12\right)\left(13x-39\right)=0\)
\(\Rightarrow\orbr{\begin{cases}4x-12=0\\13x-39=0\end{cases}\Rightarrow}\orbr{\begin{cases}4x=12\\13x=39\end{cases}\Rightarrow}\orbr{\begin{cases}x=3\\x=3\end{cases}}\)
Vậy x = 3
![](https://rs.olm.vn/images/avt/0.png?1311)
a) -12. (x-5) + 7. ( 3-x ) = 5
=> -12x + 60 + 21 - 7x = 5
=> -19x = -76
=> x= 4
b) 30. (x+2) -6. (x-5 ) -24x =100
=> 30x + 60 - 6x + 30 -24x = 100
=> 0x = 10 (vô lí )
=> ko có giá trị của x
-12.(x-5)+7.(3-x)=5
(-12).x-(-12).5+7.3-7.x=5
(-12).x-(-60)+21-7.x=5
(-12).x+60+21-7.x=5
(-12).x+(60+21)-7.x=5
(-12).x+81-7.x=5
(-12).x-7.x=5-81
-19.x=-76
x=76:(-19)
x=4
![](https://rs.olm.vn/images/avt/0.png?1311)
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 \(x=5\)
![](https://rs.olm.vn/images/avt/0.png?1311)
X - { [ -x + (x+3) ] } - [ (x+3) - (x-2)] = 0
X - { -x + x + 3 } - [ x +3 - x +2] = 0
X - 3 - 5 = 0
x - 8 = 0
x = 8
![](https://rs.olm.vn/images/avt/0.png?1311)
Đáp án là ko tồn tại x
( 2 - x ) nhân x - x + 2 =0