M=99-2:1+1=98

N=133-5:2+1=65

F=91-1:3+1=31

Q=402-2:4+1=101

So so hang: so cuoi - so dau : khoang cach +1

a) \((133-x):5=26\)

\(\implies 133-x=26\times5\)

\(\implies 133-x=130\)

\(\implies x=133-130\)

\(\implies x=3\). Vậy x=3.

b)\((3\times x-4):2-\frac{1}{4}=\frac{3}{4}\)

\(\implies (3\times x -4):2=\frac{3}{4}+\frac{1}{4}\)

\(\implies (3\times x -4):2=1\)

\(\implies 3\times x-4=2\)

\(\implies 3\times x=6\)

\(\implies x=2\). Vậy x=2.

c) \((2\times x-3)\times 3-5=10\)

\(\implies (2\times x-3)\times3 =10-5\)

\(\implies (2\times x-3)\times 3=5\)

\(\implies 2\times x-3=\frac{5}{3}\)

\(\implies 2\times x=\frac{5}{3}+3\)

\(\implies 2\times x=\frac{14}{3}\)

\(\implies x=\frac{14}{3}:2\)

\(\implies x=\frac{7}{3}\). Vậy \(x=\frac{7}{3}\)

=0

****cho mình nha bạn dễ thương

làm giúp mk bài này nhá                                                                                                              0+1+2+...+2017  có bao nhiêu số hạng

a)x³-x²=0

⇔x²(x-1)=0

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

b)3x²-5x=0

⇔ x(3x-5)=0

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{3}\end{matrix}\right.\)

c)x³=x⁵

⇔ x³(1-x²)=0

⇔ x³(1-x)(1+x)=0

⇔\(\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)

d)(2x+7)²-4(2x+7)=0

⇔ (2x+7)(2x+3)=0

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-7}{2}\\x=\dfrac{-3}{2}\end{matrix}\right.\)

a) Ta có: \(x^3-x^2=0\)

\(\Leftrightarrow x^2\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

b) Ta có: \(3x^2-5x=0\)

\(\Leftrightarrow x\left(3x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{3}\end{matrix}\right.\)

c) Ta có: \(x^3=x^5\)

\(\Leftrightarrow x^5-x^3=0\)

\(\Leftrightarrow x^3\left(x^2-1\right)=0\)

\(\Leftrightarrow x^3\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)

d) Ta có: \(\left(2x+7\right)^2-4\left(2x+7\right)=0\)

\(\Leftrightarrow\left(2x+7\right)\left(2x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-7}{2}\\x=\dfrac{-3}{2}\end{matrix}\right.\)