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Bài 2:
a: \(=248+2064-12-236\)
\(=12-12+2064=2064\)
b: \(=-298-302-300=-600-300=-900\)
c: \(=5-7+9-11+13-15=-2-2-2=-6\)
d: \(=456+58-456-38=20\)
a: \(=\dfrac{54-34}{189-119}=\dfrac{20}{70}=\dfrac{2}{7}\)
b: \(=\dfrac{6+6\cdot4+6\cdot49}{15+15\cdot4+15\cdot49}=\dfrac{6}{15}=\dfrac{2}{5}\)
c: \(=\dfrac{13\left(3-18\right)}{40\left(15-2\right)}=\dfrac{-15}{40}=-\dfrac{3}{8}\)
â/ \(-55⋮x-2\)
\(\Leftrightarrow x-2\inƯ\left(-55\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=1\\x-2=55\\x-2=-1\\x-2=-55\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=57\\x=1\\x=-53\end{matrix}\right.\)
Vậy ...........
b/ \(x^2+2x-7⋮x+2\)
Mà \(x+2⋮x+2\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+2x-7⋮x+2\\x^2+2x⋮x+2\end{matrix}\right.\)
\(\Leftrightarrow-7⋮x+2\)
\(\Leftrightarrow x+2\inƯ\left(-7\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=1\\x+2=-7\\x+2=-1\\x+2=7\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-9\\x=-3\\x=5\end{matrix}\right.\)
Vậy .........
c/ \(\left(x-15\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-15=0\\x+4=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=15\\x=-4\end{matrix}\right.\)
Vậy .........
d/ \(\left|3x-4\right|-12=13\)
\(\Leftrightarrow\left|3x-4\right|=25\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-4=25\\3x-4=-25\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{29}{3}\\x=-7\end{matrix}\right.\)
Vậy ..
1)
a) \(89-\left(73-x\right)=20\)
\(\Leftrightarrow73-x=89-20\)
\(\Leftrightarrow73-x=69\)
\(\Leftrightarrow x=73-69\)
\(\Leftrightarrow x=4\)
Vậy \(x=4\)
b) \(\left(x+7\right)-25=13\)
\(\Leftrightarrow x+7=13+25\)
\(\Leftrightarrow x+7=38\)
\(\Leftrightarrow x=38-7\)
\(\Leftrightarrow x=31\)
Vậy \(x=31\)
c) \(140:\left(x-8\right)=7\)
\(\Leftrightarrow x-8=140:7\)
\(\Leftrightarrow x-8=20\)
\(\Leftrightarrow x=20+8\)
\(\Leftrightarrow x=28\)
Vậy \(x=28\)
d) \(6x+x=5^{11}:5^9+3^1\)
\(\Leftrightarrow7x=5^{11}:5^9+3^1\)
\(\Leftrightarrow7x=5^{11-9}+3^1\)
\(\Leftrightarrow7x=5^2+3^1\)
\(\Leftrightarrow7x=25+3\)
\(\Leftrightarrow7x=28\)
\(\Leftrightarrow x=28:7\)
\(\Leftrightarrow x=4\)
Vậy \(x=4\)
e) \(4^x=64\)
\(\Leftrightarrow4^x=4^3\)
\(\Leftrightarrow x=3\)
Vậy \(x=3\)
g) \(9^{x-1}=9\)
\(\Leftrightarrow9^{x-1}=9^1\)
\(\Leftrightarrow x-1=1\)
\(\Leftrightarrow x=1+1\)
\(\Leftrightarrow x=2\)
Vậy \(x=2\)
Bài 2:Phương trình nào sau đây là phương trình bậc nhất một ẩn? Vì sao?
Phương trình bậc nhấ một ẩn là:
a) 3x + 5 = 0(vì có dạng ax+b=0)
b) 10 – 2t = 0⇔-2t+10=0 (có dạng ax+b=0)
d) 2y - 7 = 0⇔ 2y+(-7)=0 (có dạng ax+b=0)
Bài 3: Giải các phương trình sau:(phương trình tích)
a) ( 5x - 4 )( 4x + 6 ) = 0
⇔\(\left[{}\begin{matrix}5x-4=0\\4x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{4}{5}\\x=-\frac{3}{2}\end{matrix}\right.\)
b) (5x + 15)( x - 6) = 0
⇔5(x+3)(x-6)=0
⇔\(\left[{}\begin{matrix}x+3=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=6\end{matrix}\right.\)
c) ( 2x – 10)(3x + 9) = 0
⇔2(x-5)3(x+3)=0
⇔6(x-5)(x+3)=0
⇔\(\left[{}\begin{matrix}x-5=0\\x+3=0\end{matrix}\right.\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)
b) ( x - 5 )( 3 - 2x )( 3x + 4 ) = 0
⇔\(\left[{}\begin{matrix}x-5=0\\3-2x=0\\3x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\frac{3}{2}\\x=-\frac{4}{3}\end{matrix}\right.\)
c) ( 2x + 1 )( x2 + 2 ) = 0
⇔\(\left[{}\begin{matrix}2x+1=0\\x^2+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\frac{1}{2}\\x\in\varnothing\end{matrix}\right.\)
d) ( x - 2 )( 3x + 5 ) = ( x- 2)( x + 1 )
⇔( x - 2 )( 3x + 5 )-( x- 2)( x + 1 )=0
⇔( x - 2 )(3x+5-x-1)=0
⇔( x - 2 )(2x+4)=0
⇔2(x-2)(x+2)=0
⇔\(\left[{}\begin{matrix}x+2=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=2\end{matrix}\right.\)
Bài 3: Giải các phương trình sau:(phương trình tích)
a) ( 5x - 4 )( 4x + 6 ) = 0
⇔ \(\left[{}\begin{matrix}5x-4=0\\4x+6=0\end{matrix}\right.\text{⇔}\left[{}\begin{matrix}x=\frac{4}{5}\\x=\frac{-3}{2}\end{matrix}\right.\)
Vậy...
b) (5x + 15)( x - 6) = 0
\(\Leftrightarrow\left[{}\begin{matrix}5x+15=0\\x-6=0\end{matrix}\right.\text{⇔}\left[{}\begin{matrix}x=-3\\x=6\end{matrix}\right.\)
Vậy...
c) ( 2x – 10)(3x + 9) = 0
\(\Leftrightarrow\left[{}\begin{matrix}2x-10=0\\3x+9=0\end{matrix}\right.\text{⇔}\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)
Vậy...
b) ( x - 5 )( 3 - 2x )( 3x + 4 ) = 0
\(\text{⇔}\left[{}\begin{matrix}x-5=0\\3-2x=0\\3x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\frac{3}{2}\\x=\frac{-4}{3}\end{matrix}\right.\)
Vậy...
c) ( 2x + 1 )( x2 + 2 ) = 0
\(\Leftrightarrow2x+1=0\left(\text{Vì }x^2+2>0\forall x\right)\)
\(\text{⇔ }x=\frac{-1}{2}\)
Vậy..
d) ( x - 2 )( 3x + 5 ) = ( x- 2)( x + 1 )
\(\Leftrightarrow\left(x-2\right)\left(3x+5\right)-\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\2x+4=0\end{matrix}\right.\text{⇔}\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
Vậy...
a: =>25-4x=1
=>4x=24
hay x=6
b: =>2x-4=0
hay x=2
c: =>x-35=115
hay x=150
d: =>x-2=12
hay x=14
e: =>x-36=216
hay x=252