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\(\left(x-1\right)^2=\left(x-1\right)^4\)
\(\Rightarrow\left(x-1\right)^4-\left(x-1\right)^2=0\)
\(\Rightarrow\left(x-1\right)^2\left[\left(x-1\right)^2-1\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\\left(x-1\right)^2=1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x-1=1\\x-1=-1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=2\\x=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
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1. áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\frac{x+2}{3}=\frac{y-7}{5}=\frac{x+y-5}{3+5}=\frac{16}{8}=2\Rightarrow\hept{\begin{cases}x+2=6\\y-7=10\end{cases}\Leftrightarrow\hept{\begin{cases}x=4\\y=17\end{cases}}}\)
2. áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\frac{x+5}{2}=\frac{y-2}{3}=\frac{x+5-y+2}{2-3}=\frac{-10+7}{-1}=3\Rightarrow\hept{\begin{cases}x+5=6\\y-2=9\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=11\end{cases}}\)
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\(A=|x+1|+|x+2|=|-x-1|+|x+2|\)
\(\Rightarrow A\ge|-x-1+x+2|\)
\(\Rightarrow A\ge1\)
\(A=1\Leftrightarrow\hept{\begin{cases}-x-1\ge0\\x+2\ge0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x\le-1\\x\ge-2\end{cases}}\)\(\Leftrightarrow-2\le x\le-1\)
Vậy \(minA=1\Leftrightarrow-2\le x\le-1.\)
Chắc chăn đúng nha bạn
~ học tốt nha ~
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Ta có: lx-1l + l4-xl = 3 <=> lx-1l + lx-4l = 3
TH1: Nếu x < 1, ta có: TH2: Nếu 1 < x < 4, ta có: TH3: Nếu x > 4, ta có: 1 - x + 4 - x = 3 x - 1 + 4 - x = 3 x - 1 + x - 4 = 3 <=>5 - 2x = 3 <=> 3 =3 (TM) <=> 2x - 5 = 3
<=> 2x = 5 - 3 = 2 <=> x = 1;2;3;4 <=> 2x = 3 + 5 = 8 <=> x = 1 (TM) < => x = 4(TM) Vậy x = 1;2;3;4.
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$h(x)=x^2+x+1=0$
$\Rightarrow x^2+\frac{1}{2}x+\frac{1}{2}x+\frac{1}{4}+\frac{3}{4}=0$
$\Rightarrow x(x+\frac{1}{2})+\frac{1}{2}(x+\frac{1}{2})+\frac{3}{4}=0$
$\Rightarrow (x+\frac{1}{2})(x+\frac{1}{2})+\frac{3}{4}=0$
$\Rightarrow (x+\frac{1}{2})^2+\frac{3}{4}=0$
$\Rightarrow (x+\frac{1}{2})^2=\frac{-3}{4}$ (vô lí)
-Vậy: đa thức h(x) ko có nghiệm.
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\(\)
\(\frac{3}{2}-\left(x+\frac{1}{2}\right)=\frac{4}{5}\)
\(\left(x+\frac{1}{2}\right)=\frac{3}{2}-\frac{4}{5}\)
\(\left(x+\frac{1}{2}\right)=\frac{7}{10}\)
\(x=\frac{7}{10}-\frac{1}{2}\)
\(x=\frac{1}{5}\)
\(\frac{3}{2}-\left(x+\frac{1}{2}\right)=\frac{4}{5}\)
\(\left(x+\frac{1}{2}\right)=\frac{3}{2}-\frac{4}{5}\)
\(\left(x+\frac{1}{2}\right)=\frac{7}{10}\)
\(x=\frac{7}{10}-\frac{1}{2}=\frac{1}{5}\)
\(\left(x-1\right)^{x+2}=\left(x+1\right)^{x+4}\)
\(\left(x-1\right)^{x+2}=\left(x+1\right)^{x+2+2}\)
\(\left(x-1\right)^{x+2}=\left(x+1\right)^{x+2}.\left(x+1\right)^2\)
\(\left(x+1\right)^2=1\)
\(\Leftrightarrow\left(x+1\right)^2=1^2=\left(-1\right)^2\)
\(\Rightarrow\hept{\begin{cases}x+1=1\\x+1=-1\end{cases}\Rightarrow}\hept{\begin{cases}x=0\\x=-2\end{cases}}\)
cảm ơn anh nhiều ạ