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a) \(\left|4x-1\right|-\left|3x-\dfrac{1}{2}\right|=0\\ \Leftrightarrow\left|4x-1\right|=\left|3x-\dfrac{1}{2}\right|\\ \Leftrightarrow\left[{}\begin{matrix}4x-1=3x-\dfrac{1}{2}\\4x-1=\dfrac{1}{2}-3x\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}4x-3x=1-\dfrac{1}{2}\\4x+3x=\dfrac{1}{2}+1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\7x=\dfrac{3}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{3}{14}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{1}{2};\dfrac{3}{14}\right\}\) là nghiệm của pt.
b) \(\left|x-1\right|-2x=\dfrac{1}{2}\\ \Leftrightarrow\left|x-1\right|=2x+\dfrac{1}{2}\left(ĐK:x\ge\dfrac{-1}{4}\right)\\ \Leftrightarrow\left[{}\begin{matrix}x-1=2x+\dfrac{1}{2}\\x-1=-2x-\dfrac{1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x-2x=1+\dfrac{1}{2}\\x+2x=1-\dfrac{1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}-x=\dfrac{3}{2}\\3x=\dfrac{1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3}{2}\left(ktmđk\right)\\x=\dfrac{1}{6}\left(tmđk\right)\end{matrix}\right.\)
Vậy \(x=\dfrac{1}{6}\) là nghiệm của pt.
Lời giải:
a.
$|4x-1|-|3x-\frac{1}{2}|=0$
$\Leftrightarrow |4x-1|=|3x-\frac{1}{2}$
\(\Leftrightarrow \left[\begin{matrix} 4x-1=3x-\frac{1}{2}\\ 4x-1=\frac{1}{2}-3x\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{1}{2}\\ x=\frac{3}{14}\end{matrix}\right.\)
b. Nếu $x\geq 1$ thì:
$|x-1|-2x=\frac{1}{2}$
$\Leftrightarrow x-1-2x=\frac{1}{2}$
$\Leftrightarrow -x-1=\frac{1}{2}$
$\Leftrightarrow x=\frac{-3}{2}$ (vô lý vì $x\geq 1$)
Nếu $x< 1$ thì:
$1-x-2x=\frac{1}{2}$
$\Leftrightarrow x=\frac{1}{6}$ (tm)
\(a)245^2+490\cdot54+54^2-199^2\\=(245^2+2\cdot245\cdot54+54^2)-199^2\\=(245+54)^2-199^2\\=299^2-199^2\\=(299-199)(299+199)\\=100\cdot498\\=49800\\---\\b)356^2-356\cdot246+123^2-133^2\\=(356^2-2\cdot356\cdot123+123^2)-133^2\\=(356-123)^2-133^2\\=233^2-133^2\\=(233-133)(233+133)\\=100\cdot366\\=36600\)
\(---\)
\(c)468^2-412^2-110\cdot412-55^2\\=468^2-(412^2+110\cdot412+55^2)\\=468^2-(412^2+2\cdot412\cdot55+55^2)\\=468^2-(412+55)^2\\=468^2-467^2\\=(468-467)(468+467)\\=1\cdot935\\=935\\---\)
\(d)615^2+250\cdot615+125^2-540^2\\=(615^2+2\cdot615\cdot125+125^2)-540^2\\=(615+125)^2-540^2\\=740^2-540^2\\=(740-540)(740+540)\\=200\cdot1280\\=256000\)
#\(Toru\)
a) Ta có: \(M=\left(\dfrac{1}{2}x^2y\right)\cdot\left(\dfrac{2}{3}xy\right)^2\)
\(=\dfrac{1}{2}x^2y\cdot\dfrac{4}{9}x^2y^2\)
\(=\dfrac{2}{9}x^4y^3\)
b) Hệ số là \(\dfrac{2}{9}\)
Phần biến là \(x^4;y^3\)
c) Bậc là 7
d) Thay x=-1 và y=2 vào M, ta được:
\(M=\dfrac{2}{9}\cdot\left(-1\right)^4\cdot2^3=\dfrac{2}{9}\cdot8=\dfrac{16}{9}\)
a: =>7(x-5)>0
=>x-5>0
=>x>5
b: =>x-1 thuộc {1;-1;11;-11}
=>x thuộc {2;0;12;-10}
c: =>x+1+7 chia hết cho x+1
=>x+1 thuộc {1;-1;7;-7}
=>x thuộc {0;-2;6;-8}
d: =>(x+2)(x-5)<0
=>-2<x<5
\(\dfrac{5^{32}\cdot5^{43}\cdot2^{86}}{-2^{87}\cdot5^{75}}+\dfrac{5^{150}}{2}\)
\(=\dfrac{5^{150}-1}{2}\)
lớp t