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\(\left[0,\left(32\right).1,\left(5\right)-0,\left(25\right)\right].\dfrac{11}{83}\)
\(=\left[\dfrac{32}{99}.\left(1+\dfrac{5}{9}\right)-\dfrac{25}{99}\right].\dfrac{11}{83}\)
\(=\left[\dfrac{32}{99}.\dfrac{14}{9}-\dfrac{25}{99}\right].\dfrac{11}{83}\)
\(=\left[\dfrac{448}{891}-\dfrac{25}{99}\right].\dfrac{11}{83}\)
\(=\dfrac{223}{891}.\dfrac{11}{83}\)
\(=\dfrac{223}{6723}\)
a,-12(x-5)+7(3-x)=20
-12x+60+21-7x=20
-19x=-61
x=\(\frac{61}{19}\)
b,30(x+1)-3(x-5)-15x=25
30x+30+15-3x-15x=25
12x=-20
x=\(-\frac{20}{12}\)
a) \(\left(x-1\right)\left(2x-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\Rightarrow x=1\\2x-4=0\Rightarrow x=2\end{matrix}\right.\)
b) \(\left(x^2+5\right)\left(x-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x^2+5=0\Rightarrow x=-\sqrt{5}\\x-5=0\Rightarrow x=5\end{matrix}\right.\)
mà \(x\in Z\Rightarrow x=5\)
c) \(\left(x^2+5\right)\left(x^2-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x^2+5=0\Rightarrow x=-\sqrt{5}\\x^2-2=0\Rightarrow x=\sqrt{2}\end{matrix}\right.\)
mà \(x\in Z\Rightarrow x\in\varnothing\)
\(a,\left(2-x\right)\left(\dfrac{4}{5}-x\right)< 0\)
=>Trong 2 số phải có 1 số âm và 1 số dương
Mà \(2-x>\dfrac{4}{5}-x\)
=>\(\dfrac{4}{5}< x< 2\)
Vậy...
1. Ta có: \(\sqrt{23}+\sqrt{15}< \sqrt{25}+\sqrt{16}=5+4=9\)
mà \(\sqrt{83}>\sqrt{81}=9\)
\(\Rightarrow\sqrt{23}+\sqrt{15}< \sqrt{83}\)
\(\left(x-3\right)^2+\left(y+2\right)^2=0\)
\(\left\{{}\begin{matrix}\left(x-3\right)^2\ge0\forall x\\\left(y+2\right)^2\ge0\forall y\end{matrix}\right.\)
\(\Rightarrow\left(x-3\right)^2+\left(y+2\right)^2\ge0\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}\left(x-3\right)^2=0\Rightarrow x-3=0\Rightarrow x=3\\\left(y+2\right)^2=0\Rightarrow y+2=0\Rightarrow y=-2\end{matrix}\right.\)
đề sai câu b các câu sau áp dụng tương tự
c/ Vì: \(\left(x-12+y\right)^{200}+\left(x-4-x\right)^{200}=0\)
mà \(\left\{{}\begin{matrix}\left(x-12+y\right)^{200}\ge0\forall x,y\\\left(x-4-y\right)^{200}\ge0\forall x,y\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left(x-12+y\right)^{200}=0\\\left(x-4-y\right)^{200}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-12+y=0\\x-4-y=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+y=12\\x-y=4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=8\\y=4\end{matrix}\right.\)
1: =>3x+2=x+1 hoặc 3x+2=-x-1
=>2x=-1 hoặc 4x=-3
=>x=-1/2 hoặc x=-3/4
2: =>|x+2|(|x|-1|)=0
=>x=-2; x=1; x=-1
3: \(\Leftrightarrow\left\{{}\begin{matrix}x>=-1\\\left(2x+3+x+1\right)\left(2x+3-x-1\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-1\\\left(3x+4\right)\left(x+2\right)=0\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
-25-(X+5)=415+5x(X-83)
=>5x(X-83)=-25-415
=>5x(X-83)=-440
=>X-83=-440:5
=>X-83=-88
=>X=-88-(-83)
=>X=-5