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a) 3 . ( 2 - x ) + 5 . ( x - 6 ) = -98
6 - 3x + 5x - 30 = -98
6 + 2x -30 = -98
6 + 2x = -98 + 30
6 + 2x = -68
2x = -68 -6
2x = -74
x = -74 : 2
x = -37
a) \(x\left(x-6\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
b) \(\left(-7-x\right)\left(-x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}-7-x=0\\-x+5=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-7\\x=-5\end{matrix}\right.\)
c) \(\left(x+3\right)\left(x-7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x-7=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=7\end{matrix}\right.\)
d) \(\left(x-3\right)\left(x^2+12\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x^2+12=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x^2=-12\text{(vô lý)}\end{matrix}\right.\)
\(\Rightarrow x=3\)
e) \(\left(x+1\right)\left(2-x\right)\ge0\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x+1\ge0\\2-x\ge0\end{matrix}\right.\\\left[{}\begin{matrix}x+1\le0\\2-x\le0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\ge-1\\x\le2\end{matrix}\right.\\\left[{}\begin{matrix}x\le-1\\x\ge2\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}-1\le x\le2\\x\in\varnothing\end{matrix}\right.\)
\(\Rightarrow-1\le x\le2\)
f) \(\left(x-3\right)\left(x-5\right)\le0\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x-3\le0\\x-5\ge0\end{matrix}\right.\\\left[{}\begin{matrix}x-3\ge0\\x-5\le0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\le3\\x\ge5\end{matrix}\right.\\\left[{}\begin{matrix}x\ge3\\x\le5\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow3\le x\le5\)
a) =>\(\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.=>\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
b => \(\left[{}\begin{matrix}-7-x=0\\-x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-7\\x=5\end{matrix}\right.\)
d) => \(\left[{}\begin{matrix}x-3=0\\x^2+12=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x^2=-12\end{matrix}\right.\)(vô lí) => x=3
\(a,\left(-31\right).\left(x+7\right)=0\\ \Rightarrow x+7=0\\ \Rightarrow x=-7\\ b,\left(8-x\right).\left(x+13\right)=0\\ \Rightarrow\left[{}\begin{matrix}8-x=0\\x+13=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=8\\x=-13\end{matrix}\right.\\ c,\left(x^2-25\right)\left(3-x\right)=0\\ \Rightarrow\left(x-5\right)\left(x+5\right)\left(3-x\right)=0\\\Rightarrow \left[{}\begin{matrix}x-5=0\\x+5=0\\3-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=-5\\x=3\end{matrix}\right.\\ d,\left(x-3\right)\left(x^2+4\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-3=0\\x^2+4=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x^2=-4\left(loại\right)\end{matrix}\right.\\ \Rightarrow x=3\)
a. 5 - 3(x + 4) = -1
⇔ 5 - 3x - 12 = -1
⇔ 3x = -1 - 5 + 12
⇔ 3x = 6
⇔ x = 2
\(d,2x^2-3=5\)
\(\Leftrightarrow2x^2=8\)
\(\Leftrightarrow x^2=4\)
\(\Leftrightarrow x=\pm2\)
\(e,x\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=1\\x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=0\end{matrix}\right.\)
Bài 1:
a: \(x=\dfrac{2}{3}:\dfrac{3}{5}=\dfrac{2}{3}\cdot\dfrac{5}{3}=\dfrac{10}{9}\)
b: \(x=\dfrac{17}{8}:\dfrac{7}{17}=\dfrac{17}{8}\cdot\dfrac{17}{7}=\dfrac{289}{56}\)
c: \(x=-\dfrac{3}{4}:\dfrac{7}{12}=\dfrac{-3}{4}\cdot\dfrac{12}{7}=\dfrac{-63}{28}=-\dfrac{9}{4}\)
d: \(\Leftrightarrow x\cdot\dfrac{1}{6}=\dfrac{3}{8}-\dfrac{1}{4}=\dfrac{1}{4}\)
hay \(x=\dfrac{1}{4}:\dfrac{1}{6}=\dfrac{3}{2}\)
e: \(\Leftrightarrow\dfrac{1}{2}:x=-4-\dfrac{1}{3}=-\dfrac{17}{3}\)
hay \(x=-\dfrac{1}{2}:\dfrac{17}{3}=\dfrac{-3}{34}\)
a) `3(2-x)+5(x-6)=-98`
`6-3x+5x-30=-98`
`-3x+5x=-98-6+30`
`2x=-74`
`x=-37`
b) `(x+7)(8-x)=0`
\(\left[{}\begin{matrix}x+7=0\\8-x=0\end{matrix}\right.\\ \left[{}\begin{matrix}x=-7\\x=8\end{matrix}\right.\)
a, \(3\left(2-x\right)+5\left(x-6\right)=-98\)
\(6-3x+5x-30=-98\)
\(2x-24=-98\)
\(2x=-74\)
\(x=-37\)
b,\(\left(x+7\right)\left(8-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-7\\x=8\end{matrix}\right.\)
-Chúc bạn học tốt-
Bài làm
m) (x + 2).(3 - x) = 0;
=> x + 2 = 0 hoặc 3 - x = 0
=> x = -2 hoặc x = 3
Vậy x = -2 hoặc x = 3
d) 511.712 + 511.711
= 511 . ( 712 + 711 )
= 511 . [ 711 . ( 7 + 1 ) ]
= 511 . 711 . 8
= ( 5 . 7 )11 . 8
= 3511 . 8
512.712 + 9.511.711
= 511 ( 5 . 712 + 9 . 1 . 711 )
= 511 [ 711 ( 5 . 7 + 9 . 1 . 1 ) ]
= 511 ( 711 . 44 )
= 511 . 711 . 44
= 3511 . 44
m. \(\left(x+2\right)\left(3-x\right)=0\Leftrightarrow\orbr{\begin{cases}x+2=0\\3-x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=3\end{cases}}\)
d. \(\frac{5^{11}.7^{12}+5^{11}.7^{11}}{5^{12}.7^{12}+9.5^{11}.7^{11}}=\frac{5^{11}.\left(7^{12}+7^{11}\right)}{5^{11}.\left(5.7^{12}+9.7^{11}\right)}=\frac{7^{12}+7^{11}}{5.7^{12}+9.7^{11}}=\frac{1}{5.9}=\frac{1}{45}\)
q. \(\left(x-3\right)+\left(x-2\right)+\left(x-1\right)+...+10+11=11\)
\(\Rightarrow\left(x-3\right)+\left(x-2\right)+\left(x-1\right)+...+10=0\)
\(\Rightarrow\left[\left(x-3\right)+\left(x-2\right)+\left(x-1\right)\right]+(1+2+3+...+10)=0\)
\(\Rightarrow\left(x-3\right)+\left(x-2\right)+\left(x-1\right)+55=0\)
\(\Rightarrow x-3+x-2+x-1=-55\)
\(\Rightarrow3x-6=-55\)
\(\Rightarrow3x=-49\)
\(\Rightarrow x=-\frac{49}{3}\)
giải được câu b thôi.
b, (|x| + 2) . (6 - 2|x|) = 0
=> Với mọi số nguyên x thuộc Z thì |x| > 2
=> 6 - 2|x| = 0
=> 2|x| = 6 - 0
=> 2|x| = 6
=> |x| = 6 / 3
=> |x| = 2 và x = 2 hoặc x = -2