2.(x-3) + 3 . (x-13) = 0
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a/ \(x\left(x+7\right)=0\) \(\Rightarrow\left[\begin{matrix}x=0\\x+7=0\end{matrix}\right.\) \(\Rightarrow\left[\begin{matrix}x=0\\x=-7\end{matrix}\right.\)
b/ \(\left(-x+5\right)\left(3-x\right)=0\) \(\Rightarrow\left[\begin{matrix}-x+5=0\\3-x=0\end{matrix}\right.\) \(\Rightarrow\left[\begin{matrix}x=5\\x=3\end{matrix}\right.\)
c/ \(\left|x-1\right|=3\) \(\Rightarrow\left[\begin{matrix}x-1=3\\1-x=3\end{matrix}\right.\) \(\Rightarrow\left[\begin{matrix}x=4\\x=-2\end{matrix}\right.\)
d/ \(-13\left|x\right|=-26\) \(\Rightarrow\left|x\right|=2\) \(\Rightarrow x=\pm2\)
e/ \(x.x-8=-2.\left(-13\right)-\left(-2\right)\)
\(\Rightarrow x^2=36\) \(\Rightarrow\left|x\right|=6\) \(\Rightarrow x=\pm6\)
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1.\(\left(x-40\right)\left(x-7\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-40=0\\x-7=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=40\\x=7\end{cases}}\)
2. \(\left(x-12\right)\left(x+13\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-12=0\\x+13=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=12\\x=-13\end{cases}}\)
3. \(\left(2-x-4\right)\left(3x-9\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2-x-4=0\\3x-9=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-2\\x=3\end{cases}}\)
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a. 3.(x-2)+2.(x-3)=13
x=5
b. (x+1).(2-x)-(3x+5).(x+2)=-4x2+1
x=-9/10
c.x.(5-2x)+2x.(x-1)=13
x=13/3
d. (2x+3)2-(x-1)2=0
x=-2/3
e. x2.(3x-2)-8+12=0
x vô ngiệm
f x2+x=0
x=-1
g. x3-5x=0
x=0
~~~~~~~~~~~ai đi ngang qua nhớ để lại k ~~~~~~~~~~~~~
~~~~~~~~~~~~ Chúc bạn sớm kiếm được nhiều điểm hỏi đáp ~~~~~~~~~~~~~~~~~~~
a) \(3\left(x-2\right)+2\left(x-3\right)=1\)\(3\)
\(3x-6+2x-6=13\)
\(5x=13+6+6\)
\(5x=25\)
\(x=25\)
c) \(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(5x-2x^2+2x^2-2x=13\)
\(3x=13\)
\(x=\frac{13}{3}\)
d) \(\left(2x+3\right)^2-\left(x-1\right)^2=0\)
\(\left(2x+3-x+1\right)\left(2x+3+x-1\right)=0\)
\(\left(x+4\right)\left(3x+2\right)=0\)
\(\orbr{\begin{cases}x+4=0\\3x+2=0\end{cases}}=>\orbr{\begin{cases}x=-4\\x=\frac{-2}{3}\end{cases}}\)
f) \(x^2+x=0\)
\(x\left(x+1\right)=0\)
\(=>\orbr{\begin{cases}x=0\\x+1=0\end{cases}=>\orbr{\begin{cases}x=0\\x=-1\end{cases}}}\)
g) \(x^3-5x=0\)
\(x^2\left(x-5\right)=0\)
\(=>\orbr{\begin{cases}x^2=0\\x-5=0\end{cases}}\)
\(=>\orbr{\begin{cases}x=0\\x=5\end{cases}}\) \(\)
\(\)
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a: =16-2+91=14+91=105
b: =9*5+8*10-27=45+53=98
c: =32+65-3*8=8+65=73
d; \(=5^3-10^2=125-100=25\)
e: \(=4^2-3^2+1=8\)
f: =9*16-16*8-8+16*4
=16(9-8+4)-8
=16*5-8
=72
a) \(2^4-50:25+13\cdot7\)
\(=2^4-2+91\)
\(=16-2+91\)
\(=14+91\)
\(=105\)
b) \(3^2\cdot5+2^3\cdot10-3^4:3\)
\(=9\cdot5+8\cdot10-3^3\)
\(=45+80-27\)
\(=98\)
c) \(2^5+5\cdot13-3\cdot2^3\)
\(=32+65-3\cdot8\)
\(=32+65-24\)
\(=73\)
d) \(5^{13}:5^{10}-5^2\cdot2^2\)
\(=5^{13-10}-\left(5\cdot2\right)^2\)
\(=5^3-10^2\)
\(=125-100\)
\(=25\)
e) \(4^5:4^3-3^9:3^7+5^0\)
\(=4^{5-3}-3^{9-7}+1\)
\(=4^2-3^2+1\)
\(=16-9+1\)
\(=8\)
f) \(3^2\cdot2^4-2^3\cdot4^2-2^3\cdot5^0+4^2\cdot2^2\)
\(=3^2\cdot4^2-2^3\cdot4^2-2^3\cdot1+4^2\cdot2^2\)
\(=4^2\cdot\left(3^2-2^3+2^2\right)-2^3\)
\(=4^2\cdot\left(9-8+4\right)-8\)
\(=16\cdot5-8\)
\(=72\)
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a) Ta có : ( x + 1 ).( 3 - x ) > 0
Th1 : \(\hept{\begin{cases}x+1>0\\3-x>0\end{cases}\Rightarrow\hept{\begin{cases}x>-1\\x>3\end{cases}\Rightarrow}x>3}\)
Th2 : \(\hept{\begin{cases}x+1< 0\\3-x< 0\end{cases}\Rightarrow\hept{\begin{cases}x< -1\\x< 3\end{cases}\Rightarrow}x< -1}\)
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1) Do x ∈ Z và 0 < x < 3
⇒ x ∈ {1; 2}
2) Do x ∈ Z và 0 < x ≤ 3
⇒ x ∈ {1; 2; 3}
3) Do x ∈ Z và -1 < x ≤ 4
⇒ x ∈ {0; 1; 2; 3; 4}
vì tổng các số hạng bằng 0
=> các số hạng bằng 0
2.(x-3)=0
x-3=0
x=3
3.(x-13)=0
x-13 =0
x=13
vậy \(x_1\)=3
\(x_2\)=13