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Ta có: \(\hept{\begin{cases}\left(x+20\right)^{100}\ge0;\forall x,y\\\left|y+4\right|\ge0;\forall x,y\end{cases}}\)\(\Rightarrow\left(x+20\right)^{100}+\left|y+4\right|\ge0;\forall x,y\)
Do đó \(\left(x+20\right)^{100}+\left|y+4\right|=0\)
\(\Leftrightarrow\hept{\begin{cases}\left(x+20\right)^{100}=0\\\left|y+4\right|=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=-20\\y=-4\end{cases}}\)
Vậy ...
ta có (x+20)^100 \(\ge\) 0 (vì 100 là số mũ chẵn)
\(|y+4|\ge0\)
=>\(\left(x+20\right)^{100}+|y+4|\ge0\)
dấu "=+ xảy ra khi x=-20 và y=-4
\(4x-3-x-5-=x+2-2x+2.10\)
\(3x-8=x+2-2x+20\)
\(3x=x+10-2x+20\)
\(3x=-x+30\)
\(4x=30\)
\(x=\frac{30}{4}=7,5\)
\(2\left(x+1\right)-3\left(x+2\right)=10\Leftrightarrow2x+2-3x-6=10\)
\(\Leftrightarrow-x-4=10\Leftrightarrow x=-4-10=-14\) vậy \(x=-14\)
=>
\(\frac{x+2}{10^{10}}+\frac{x+2}{11^{11}}-\frac{x+2}{12^{12}}-\frac{x+2}{13^{13}}=0\)
=>\(\left(x+2\right).\left(\frac{1}{10^{10}}+\frac{1}{11^{11}}-\frac{1}{12^{12}}-\frac{1}{13^{13}}=0\right)\)
vì \(\frac{1}{10^{10}}>\frac{1}{11^{11}}>\frac{1}{12^{12}}>\frac{1}{13^{13}}=>\frac{1}{10^{10}}+\frac{1}{11^{11}}-\frac{1}{12^{12}}-\frac{1}{13^{13}}\ne0\)
=>x+2=0
=>x=-2
tick nhé
\(\frac{x+2}{10^{10}}+\frac{x+2}{11^{11}}=\frac{x+2}{12^{12}}+\frac{x+2}{13^{13}}\)
\(=>\frac{x+2}{10^{10}}+\frac{x+2}{11^{11}}-\frac{x+2}{12^{12}}-\frac{x+2}{13^{13}}=0\)
\(=>\left(x+2\right).\left(\frac{1}{10^{10}}+\frac{1}{11^{11}}-\frac{1}{12^{12}}-\frac{1}{13^{13}}\right)=0\)
Vì \(\frac{1}{10^{10}}>\frac{1}{11^{11}}>\frac{1}{12^{12}}>\frac{1}{13^{13}}\)
Nên \(\frac{1}{10^{10}}+\frac{1}{11^{11}}-\frac{1}{12^{12}}-\frac{1}{13^{13}}\)khác 0
Suy ra: x+2=0
x =0 - 2
x = -2
TICK NHA
\(2(x-5)-3(x+6)=10+x\)
\(\Rightarrow2x-10-3x+18=10x\)
\(\Rightarrow2x-10-3x-18=10+x\)
\(\Rightarrow-x-28=10+x\)
\(\Rightarrow-x-x=10+28\)
\(\Rightarrow-2x=38\)
\(\Rightarrow x=-19\)
Vậy ..
\(2\left(x-5\right)-\left(3x+6\right)=10+x\)
\(\Rightarrow2x-10-3x-18=10+x\)
\(\Rightarrow-x-28=10+x\)
\(\Rightarrow-x=38+x\)
\(\Rightarrow-2x=38\Leftrightarrow x=-19\)
Minh biet dap an nhung trinh bay the nao thi minh bo tay! Cac ban giup minh nhe
<=> x^20 - x^10 = 0
<=> x^10.x^10 - x^10 = 0
<=> x^10( x^10 - 1 ) = 0
=> x^10 = 0 hoac x^10 - 1 = 0
=> x = 0 hoac x = -1;1