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\(\Leftrightarrow\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
câu a giống Võ Đoan Nhi
câu b:
( x2 + 2x -11 ) : ( x + 2)
=> x2 + 2x -11 : ( x + 2)
=> x(x+2) -11 : ( x + 2)
Vì x( x + 2) : ( x + 2) nên -11 : ( x + 2)
=> x + 2 thuộc ước của -11
ta lập bảng..............
\(3x+4⋮x-3\)
\(\Leftrightarrow3\left(x-3\right)+10\)\(⋮x-3\)
-Mà: \(3\left(x-3\right)⋮x-3\Rightarrow10⋮x-3\)
\(\Leftrightarrow x-3\inƯ\left(10\right)\Leftrightarrow x-3\in\left\{\pm1;\pm2;\pm5;\pm10\right\}\)
-Lập bảng:.....
a/ Ta có :
\(\left|x-5\right|\ge0\forall x\)
\(\Leftrightarrow\left|x-5\right|+3\ge3\forall x\)
\(\Leftrightarrow A\ge3\)
Dấu "=" xảy ra khi : \(\left|x-5\right|=0\)
\(\Leftrightarrow x=5\)
Vậy \(A_{Min}=3\Leftrightarrow x=5\)
b,c tương tự
Dùng công thức tính tổng
\(1+2+3+...+x=11325\)
\(\Leftrightarrow\frac{x\left(x+1\right)}{2}=11325\)
\(\Leftrightarrow x\left(x+1\right)=22650\)
\(\Leftrightarrow x\left(x+1\right)=150.151\)
Nên x = 150
Vậy ,,,
\(1+2+3+4+...+x=11325\)
\(\Rightarrow\frac{x\left(x+1\right)}{2}=11325\)
\(\Rightarrow x\left(x+1\right)=11325\times2\)
\(\Rightarrow x\left(x+1\right)=22650\)
\(\Rightarrow150\times151=22650\)
\(\Rightarrow x=150\)
1, Ta có :
\(x+\frac{3}{5}=\frac{4}{7}\div\frac{8}{21}\)
\(x+\frac{3}{5}=\frac{4}{7}\times\frac{21}{8}\)
\(x+\frac{3}{5}=\frac{3}{2}\)
\(x=\frac{3}{2}-\frac{3}{5}\)
\(x=\frac{15}{10}-\frac{6}{10}\)
\(x=\frac{9}{10}\)
Vậy x = \(\frac{9}{10}\)
2, Ta có :
\(\frac{2}{3}+\frac{3}{4}\div x=-\frac{1}{6}\)
\(\frac{3}{4}\div x=-\frac{1}{6}-\frac{2}{3}\)
\(\frac{3}{4}\div x=-\frac{1}{6}-\frac{4}{6}\)
\(\frac{3}{4}\div x=-\frac{5}{6}\)
\(x=\frac{3}{4}\div\left(-\frac{5}{6}\right)\)
\(x=\frac{3}{4}\times\left(-\frac{6}{5}\right)\)
\(x=-\frac{9}{10}\)
Vậy x = \(-\frac{9}{10}\)
a) \(\frac{3}{4}x-\frac{1}{4}=2\left(x-3\right)+\frac{1}{4}x\)
\(\frac{3}{4}x-\frac{1}{4}=2x-6+\frac{1}{4}x\)
\(\frac{3}{4}x-2x-\frac{1}{4}x=\frac{1}{4}-6\)
\(x\left(\frac{3}{4}-2-\frac{1}{4}\right)=-\frac{23}{4}\)
\(-\frac{3}{2}x=-\frac{23}{4}\)
\(x=-\frac{23}{4}\div\left(-\frac{3}{2}\right)\)
\(x=\frac{23}{6}\)
Đặt \(\left(1+5+5^2+5^3+...+5^{2010}+5^{2011}\right)\) là A
\(\Rightarrow5A=5+5^2+5^3+5^4+...+5^{2011}+5^{2012}\)
\(\Rightarrow5A-A=5+5^2+5^3+5^4+...+5^{2011}+5^{2012}-1-5-5^2-5^3-...-5^{2010}-5^{2011}\)
\(\Rightarrow4A=5^{2012}-1\)
\(\Rightarrow A=\frac{1}{4}\left(5^{2012}-1\right)\)
Thay A vào, ta có:
\(\frac{1}{4}\left(5^{2012}-1\right)\left(x-1\right)=5^{2012}-1\)
\(\frac{1}{4}\left(x-1\right)=1\)
\(x-1=4\)
\(x=3\)
=> 20 - 4x - 3 - 3x = - 9
=> 17 - x = - 9
=> x = 26
Vậy x = 26
K mk nha
\(4.\left(5-x\right)-3.\left(1-x\right)=-3^2\)
\(20-4x-3-3x=-9\)
\(17-x=-9\)
\(x=17+9\)
\(x=26\)