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a) \(\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
b) \(\left(x^2+5\right)\left(x^2-25\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+5=0\\x^2-25=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2=-5\\x^2=25\end{matrix}\right.\) \(\Leftrightarrow x^2=25\) \(\Leftrightarrow x=\pm5\)
Bài 2:
a: =>x=0 hoặc x=-3
b: =>x-2=0 hoặc 5-x=0
=>x=2 hoặc x=5
c: =>x-1=0
hay x=1
\(a,\dfrac{-5}{x-3}< 0\Leftrightarrow x-3>0\left(-5< 0\right)\Leftrightarrow x>3\\ b,\dfrac{3-x}{x^2+1}\ge0\Leftrightarrow3-x\ge0\left(x^2+1>0\right)\Leftrightarrow x\le3\\ c,\dfrac{\left(x-1\right)^2}{x-2}< 0\Leftrightarrow x-2< 0\left[\left(x-1\right)^2\ge0\right]\Leftrightarrow x< 2\)
a) \(\left|3x-\dfrac{1}{2}\right|+\left|\dfrac{1}{4}y+\dfrac{3}{5}\right|=0\)
Do \(\left|3x-\dfrac{1}{2}\right|,\left|\dfrac{1}{4}y+\dfrac{3}{5}\right|\ge0\forall x,y\)
\(\Rightarrow\left\{{}\begin{matrix}3x-\dfrac{1}{2}=0\\\dfrac{1}{4}y+\dfrac{3}{5}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{6}\\y=-\dfrac{12}{5}\end{matrix}\right.\)
b) \(\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|+\left|\dfrac{5}{7}y-\dfrac{1}{2}\right|\le0\)
Do \(\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|,\left|\dfrac{5}{7}y-\dfrac{1}{2}\right|\ge0\forall x,y\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3}{2}x+\dfrac{1}{9}=0\\\dfrac{5}{7}y-\dfrac{1}{2}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{2}{27}\\y=\dfrac{7}{10}\end{matrix}\right.\)
a: =>x-2=0 và y+3=0
=>x=2 và y=-3
b: =>|x-2|=|x+3|
=>x-2=x+3 hoặc x+3=2-x
=>2x=-1
=>x=-1/2
c: TH1: x<-5/4
Pt sẽ là -x-5/4+3/4-x=1
=>-2x-1/2=1
=>-2x=3/2
=>x=-3/4(loại)
TH2: -5/4<=x<3/4
Pt sẽ là x+5/4+3/4-x=1
=>8/4=1(loại)
TH3: x>=3/4
Pt sẽ là x-3/4+x+5/4=1
=>2x+1/2=1
=>2x=1/2
=>x=1/4(loại)
bạn đăg tách ra cho m.n cùng giúp nhé
Bài 2 :
a, \(A=\left|2x-4\right|+2\ge2\)
Dấu ''='' xảy ra khi x = 2
Vậy GTNN A là 2 khi x = 2
b, \(B=\left|x+2\right|-3\ge-3\)
Dấu ''='' xảy ra khi x = -2
Vậy GTNN B là -3 khi x = -2
a: Ta có: \(\left(\dfrac{3}{2}x-\dfrac{1}{5}\right)^2\cdot\left(x^2+\dfrac{1}{2}\right)=0\)
\(\Leftrightarrow x\cdot\dfrac{3}{2}=\dfrac{1}{5}\)
hay \(x=\dfrac{1}{5}:\dfrac{3}{2}=\dfrac{2}{15}\)
b: Ta có: \(\dfrac{x+1}{99}+\dfrac{x+2}{98}+\dfrac{x+3}{97}+\dfrac{x+4}{96}=-4\)
\(\Leftrightarrow x+100=0\)
hay x=-100
a: \(\left(\dfrac{1}{4}-x\right)\left(x+\dfrac{2}{5}\right)=0\)
=>\(\left[{}\begin{matrix}\dfrac{1}{4}-x=0\\x+\dfrac{2}{5}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=-\dfrac{2}{5}\end{matrix}\right.\)
b: \(\left|2x+1\right|+\dfrac{3}{2}=2\)
=>\(\left|2x+1\right|=\dfrac{1}{2}\)
=>\(\left[{}\begin{matrix}2x+1=\dfrac{1}{2}\\2x+1=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-\dfrac{1}{2}\\2x=-\dfrac{3}{2}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-\dfrac{1}{4}\\x=-\dfrac{3}{4}\end{matrix}\right.\)
c: (2x-3)2=36
=>\(\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=9\\2x=-3\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
d: \(7^{x+2}+2\cdot7^x=357\)
=>\(7^x\cdot49+7^x\cdot2=357\)
=>\(7^x=7\)
=>x=1
a) \(\left(\dfrac{1}{4}-x\right)\left(x+\dfrac{2}{5}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{4}-x=0\\x+\dfrac{2}{5}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=-\dfrac{2}{5}\end{matrix}\right.\)
\(---\)
b) \(\left|2x+1\right| +\dfrac{2}{3}=2\)
\( \Rightarrow\left|2x+1\right|=2-\dfrac{2}{3}\)
\(\Rightarrow\left|2x+1\right|=\dfrac{4}{3}\)
\(\Rightarrow\left[{}\begin{matrix}2x+1=\dfrac{4}{3}\\2x+1=-\dfrac{4}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=\dfrac{1}{3}\\2x=-\dfrac{7}{3}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\\x=-\dfrac{7}{6}\end{matrix}\right.\)
\(---\)
c) \(\left(2x-3\right)^2=36\)
\(\Rightarrow\left(2x-3\right)^2=\left(\pm6\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=9\\2x=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
\(---\)
d) \(7^{x+2}+2\cdot7^x=357\)
\(\Rightarrow7^x\cdot7^2+2\cdot7^x=357\)
\(\Rightarrow7^x\cdot\left(7^2+2\right)=357\)
\(\Rightarrow7^x\cdot\left(49+2\right)=357\)
\(\Rightarrow7^x\cdot51=357\)
\(\Rightarrow7^x=357:51\)
\(\Rightarrow7^x=7\)
\(\Rightarrow x=1\)
a) Vì |x| và |x+2| luôn lớn hơn hoặc bằng 0
mà |x| + |x+2| = 0
=> \(\hept{\begin{cases}x=0\\x+2=0\end{cases}}\)
=> \(\hept{\begin{cases}x=0\\x=-2\end{cases}}\)
Vậy,.........
a) Nhận xét : \(\left|x\right|\ge0;\left|x+2\right|\ge0\Rightarrow\left|x\right|+\left|x+2\right|\ge0\)
Dấu "=" xảy ra khi : | x | = 0 và | x + 2 | = 0
\(\hept{\begin{cases}\left|x\right|=0\\\left|x+2\right|=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\x=-2\end{cases}}}\)
=> Không có x thỏa mãn
b) \(\left|x\left(x^2-\frac{5}{4}\right)\right|=x\)
Th1:\(\)
\(x.\left(x^2-\frac{5}{4}\right)=-x\)
\(\Leftrightarrow x^2-\frac{5}{4}=-1\)
\(\Leftrightarrow x^2=\frac{1}{4}\)
\(\Leftrightarrow x=\frac{1}{2}\)
\(x.\left(x^2-\frac{5}{4}\right)=x\)
\(\Leftrightarrow x^2-\frac{5}{4}=1\)
\(\Leftrightarrow x^2=\frac{9}{4}\)
\(\Leftrightarrow x=\frac{3}{2}\)