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1. \(\Rightarrow x-\frac{1}{2}=0\Rightarrow x=\frac{1}{2}\)
2. \(\Rightarrow x-2=1\Rightarrow x=3\)
hoặc \(x-2=-1\Rightarrow x=1\)
3. \(\Rightarrow2x-1=-8\Rightarrow x=\frac{-9}{2}\)
4. \(\Rightarrow x+\frac{1}{2}=\frac{1}{4}\Rightarrow x=\frac{1}{4}\)
hoặc \(x+\frac{1}{2}=\frac{-1}{4}\Rightarrow x=\frac{-3}{4}\)
a) Thiếu đề (hoặc sai)
b) x đâu?
c)\(3x-1=x+2\)
\(\Rightarrow3x-x=2+1\)
\(\Rightarrow2x=3\)
\(\Rightarrow x=\frac{3}{2}\)
c) \(\frac{x+2}{5}=\frac{2-3x}{3}\)
\(\Rightarrow3.\left(x+2\right)=5.\left(2-3x\right)\)
\(\Rightarrow3x+6=10-15x\)
\(\Rightarrow3x+15x=10-6\)
\(\Rightarrow18x=4\)
\(\Rightarrow x=\frac{4}{18}=\frac{2}{9}\)
câu 1 là \(x\times\left(4.6+\frac{3}{5}\right)=7.2-8.15\)
câu 2 là \(42+\frac{3}{7}.\left[3\times x-1=12\right]\)
Bài 1:
\(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{6}\right|+...+\left|x+\frac{1}{101}\right|=101x\)
Ta thấy:
\(VT\ge0\Rightarrow VP\ge0\Rightarrow101x\ge0\Rightarrow x\ge0\)
\(\Rightarrow\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{6}\right)+...+\left(x+\frac{1}{101}\right)=101x\)
\(\Rightarrow\left(x+x+...+x\right)+\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{101}\right)=0\)
\(\Rightarrow10x+\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{10.11}\right)=0\)
\(\Rightarrow10x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\right)=0\)
\(\Rightarrow10x+\left(1-\frac{1}{11}\right)=0\)
\(\Rightarrow10x+\frac{10}{11}=0\)
\(\Rightarrow10x=-\frac{10}{11}\Rightarrow x=-\frac{1}{11}\)(loại,vì x\(\ge\)0)
Bài 2:
Ta thấy: \(\begin{cases}\left(2x+1\right)^{2008}\ge0\\\left(y-\frac{2}{5}\right)^{2008}\ge0\\\left|x+y+z\right|\ge0\end{cases}\)
\(\Rightarrow\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|\ge0\)
Mà \(\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|=0\)
\(\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|=0\)
\(\Rightarrow\begin{cases}\left(2x+1\right)^{2008}=0\\\left(y-\frac{2}{5}\right)^{2008}=0\\\left|x+y+z\right|=0\end{cases}\)\(\Rightarrow\begin{cases}2x+1=0\\y-\frac{2}{5}=0\\x+y+z=0\end{cases}\)
\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\x+y+z=0\end{cases}\)\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\-\frac{1}{2}+\frac{2}{5}+z=0\end{cases}\)
\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\-\frac{1}{10}=-z\end{cases}\)\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\z=\frac{1}{10}\end{cases}\)
1. tự sắp nha
2. * P(x) + Q(x) = 3x5 + 5x - 4x4 - 2x3 +6 + 4x2 + 2x4 - x + 3x2 - 2x3 + 1/4 - x5
= 3x5 - x5 + (-4x4 + 2x4) + 2x3 - 2x3 + 4x2 + 3x2 + 5x - x + 6 + 1/4
= 2x5 - 2x4 + 7x2 + 4x + 25/4
* P(x) - Q(x) = (3x5 + 5x - 4x4 - 2x3 + 6 + 4x2) - (2x4 - x + 3x2 - 2x3 + 1/4 - x5)
= 3x5 + 5x - 4x4 - 2x3 + 6 + 4x2 - 2x4 + x - 3x2 + 2x3 - 1/4 + x5
= 3x5 + x5 + (-4x4 - 2x4) + (-2x3 + 2x3) + 4x2 - 3x2 + 5x + x + 6 - 1/4
= 4x5 - 6x4 + x2 + 6x + 23/4
3. ko bít
\(\text{1. Ta có hai trường hợp :}\)
\(\text{ TH1 : 3x = 0}=>x=0.\)
\(\text{ TH2 : x -}\frac{1}{2}=2=>x=\frac{5}{2}.\)
\(\text{Vậy x = 0 , x = }\frac{5}{2}.\)
\(1,\)\(3x\left(x-\frac{1}{2}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x-\frac{1}{2}=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{2}\end{cases}}}\)
\(2,\)\(\left(\frac{4}{3}+\frac{2}{5}\right)-x=\frac{1}{2}+\frac{1}{3}\)
\(\Rightarrow x=\frac{4}{3}+\frac{2}{5}-\frac{1}{2}-\frac{1}{3}\)
\(\Rightarrow x=1+\frac{4}{10}-\frac{5}{10}=1-\frac{1}{10}=\frac{9}{10}\)
\(\left(2x-3\right)^3=\left(1-x\right)^3\)
\(=>2x-3=1-x\)
\(=>3x=4=>x=\frac{4}{3}\)
\(\left(x-1\right)^3-\left(x-1\right)^2=0\)
\(\left(x-1\right)^2.\left[\left(x-1\right)-1\right]=0\)
\(=>\orbr{\begin{cases}\left(x-1\right)^2=0\\x-2=0\end{cases}}\)
\(=>\orbr{\begin{cases}x=1\\x=2\end{cases}}\)
Vậy..