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\(\left|a+2\right|=a\)
\(\Rightarrow a+2=\hept{\begin{cases}a\\-a\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}a-a=2\\-a-a=2\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}0=2\left(loai\right)\\-2a=2\end{cases}}\)
\(\Rightarrow a=-1\)
a,ta có:\(a+b+c=0\Rightarrow\)\(a+b=-c;a+c=-b;b+c=-a\)
\(\Rightarrow A=\left(\frac{a+b}{b}\right)\left(\frac{b+c}{c}\right)\left(\frac{c+a}{a}\right)=-1\)
b)
*Ta thấy x = 4 thì ta có (4 – 4).f(4) = (4– 5).f(4 + 2) suy ra f(6) = 0 hay x = 6 là nghiệm của f(x)
* Với x = 5 thì ta có (5 – 4).f(5) = (5– 5).f(5 + 2)suy ra f(5) = 0 hay x = 5 là nghiệm của f(x)
Vậy f(x) có ít nhất hai nghiệm.
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a)\(|2x-3|+3=16\)
\(|2x-3|=16-3\)
\(|2x-3|=13\)
\(\Leftrightarrow\orbr{\begin{cases}2x-3=13\\2x-3=-13\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=13+3\\2x=-13+3\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=16\\2x=-10\end{cases}}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=16:2\\x=-10:2\end{cases}\Leftrightarrow\orbr{\begin{cases}x=8\\x=-5\end{cases}}}\)
Vây\(x\in\left\{8;-5\right\}\)
c)\(\left(x-\frac{1}{2}\right)\times\left(1+5\right)=0\)
\(\left(x-\frac{1}{2}\right)\times6=0\)
\(x-\frac{1}{2}=0:6\)
\(x-\frac{1}{2}=0\)
\(x=0+\frac{1}{2}\)
\(x=\frac{1}{2}\)
a)
<=> \(x\left(0,2-1,2\right)+3,7=-6,3\)
<=> \(-x=-10\)
<=> \(x=10\)
b)
<=> \(x\left(x-1\right)=0\)
<=> \(\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
d)
<=> \(2\sqrt{x+1}=8\)
<=> \(\sqrt{x+1}=4\)
<=> \(x=15\)
e)
<=> \(\orbr{\begin{cases}1-x=\sqrt{2}-0,\left(1\right)\\1-x=0,\left(1\right)-\sqrt{2}\end{cases}}\)
<=> \(\orbr{\begin{cases}1+0,\left(1\right)-\sqrt{2}=x\\x=1+\sqrt{2}-0,\left(1\right)\end{cases}}\)
a) 0,2x + ( -1, 2 )x + 3, 7 = -6, 3
<=> x( 0,2 - 1, 2 ) + 3, 7 = -6, 3
<=> -x = -10
<=> x = 10
b) x2 = x
<=> x2 - x = 0
<=> x( x - 1 ) = 0
<=> \(\orbr{\begin{cases}x=0\\x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
c) 0,(12) : 1,(6) = x : 0,(4)
<=> 4/33 : 5/3 = x : 4/9
<=> 4/55 = x : 4/9
<=> x = 16/495
d) \(2\sqrt{x+1}-3=5\)
\(\Leftrightarrow2\sqrt{x+1}=8\)
\(\Leftrightarrow\sqrt{x+1}=4\)
\(\Leftrightarrow x+1=16\)
\(\Leftrightarrow x=15\)
e) \(\left|1-x\right|=\sqrt{2}-0,\left(1\right)\)
\(\Leftrightarrow\left|1-x\right|=\sqrt{2}-\frac{1}{9}\)
\(\Leftrightarrow\left|1-x\right|=\frac{-1+9\sqrt{2}}{9}\)
\(\Leftrightarrow\orbr{\begin{cases}1-x=\frac{-1+9\sqrt{2}}{9}\\1-x=\frac{1-9\sqrt{2}}{9}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{10-9\sqrt{2}}{9}\\x=\frac{8+9\sqrt{2}}{9}\end{cases}}\)
a) ta có:
\(|2x-6|+5x=9\Leftrightarrow|2x-6|=9-5x\)
\(2x-6=9-5x\Leftrightarrow7x=15\Leftrightarrow x=\frac{15}{7}\)
\(2x-6=5x-9\Leftrightarrow3x=3\Leftrightarrow x=1\)
b) Ta có:
\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+329}{5}+4=0\)
\(\Leftrightarrow\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)
do \(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\ne0\)nên \(x+329=0\Leftrightarrow x=-329\)
Vậy ............................................. chúc bn hok tốt ^-^
a) \(\left(x-\frac{2}{5}\right).\left(x+\frac{3}{7}\right)<0\)
\(\Rightarrow x-\frac{2}{5}<0\) hoặc \(x-\frac{2}{5}>0\)
\(x+\frac{3}{7}>0\) \(x+\frac{3}{7}<0\)
\(\Rightarrow x<\frac{2}{5}\) hoặc \(x>\frac{2}{5}\)
\(x>-\frac{3}{7}\) \(x<-\frac{3}{7}\)
\(\Rightarrow-\frac{3}{7} hoặc \(x\in rỗng\)
vậy \(-\frac{3}{7}
b) \(\frac{1}{2}-\left(\frac{1}{3}+\frac{1}{4}\right)\le x\le\frac{1}{24}-\left(\frac{1}{8}-\frac{1}{3}\right)\)
\(\frac{-1}{12}\le x\le\frac{1}{4}\)
\(\frac{-1}{12}\le x\le\frac{3}{12}\)
\(\Rightarrow x=\frac{-1}{12};0;\frac{1}{12};\frac{2}{12};\frac{3}{12}\)
a) \(\left|x-\frac{2}{5}\right|-\frac{1}{4}=0\)
=> \(\left|x-\frac{2}{5}\right|=\frac{1}{4}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{2}{5}=\frac{1}{4}\\x-\frac{2}{5}=-\frac{1}{4}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{13}{20}\\x=\frac{3}{20}\end{cases}}\)
Vậy
b) \(\left|x+0,8\right|-2,9=-12\)
\(\Rightarrow\left|x+0,8\right|=-12+2,9\)
\(\left|x+0,8\right|=-9,1\)
\(\Rightarrow\orbr{\begin{cases}x+0,8=9,1\\x+0,8=-9,1\end{cases}}\Rightarrow\orbr{\begin{cases}x=8,3\\x=-9,9\end{cases}}\)
Vậy ...
c) |x-0,987|+6,2=-3
|x-0,987|=-3-6,2
|x-0,987|=-9,2
\(\Rightarrow\orbr{\begin{cases}x-0,987=-9,2\\x-0,987=9,2\end{cases}}\Rightarrow\orbr{\begin{cases}x=-8,213\\10,187\end{cases}}\)
Vậy ...