Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) Ta có: \(3-\left(17-x\right)=-12\)
\(\Leftrightarrow3-17+x+12=0\)
\(\Leftrightarrow x-2=0\)
hay x=2
Vậy: x=2
b) Ta có: \(\left(2x+4\right)\left(10-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+4=0\\10-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-4\\2x=10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=5\end{matrix}\right.\)
Vậy: \(x\in\left\{-2;5\right\}\)c) Ta có: \(\left|x-9\right|=-2+17\)
\(\Leftrightarrow\left|x-9\right|=15\)
\(\Leftrightarrow\left[{}\begin{matrix}x-9=15\\x-9=-15\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=24\\x=-6\end{matrix}\right.\)
Vậy: \(x\in\left\{24;-6\right\}\)
a) \(\dfrac{13}{20}+\dfrac{3}{5}+x=\dfrac{5}{6}\)
\(\Rightarrow\dfrac{5}{4}+x=\dfrac{5}{6}\)
\(\Rightarrow x=\dfrac{5}{6}-\dfrac{5}{4}\)
\(\Rightarrow x=\dfrac{-5}{12}\)
b) \(x+\dfrac{1}{3}=\dfrac{2}{5}-\dfrac{-1}{3}\)
\(\Rightarrow x+\dfrac{1}{3}=\dfrac{11}{15}\)
\(\Rightarrow x=\dfrac{11}{15}-\dfrac{1}{3}\)
\(\Rightarrow x=\dfrac{2}{5}\)
c)\(\dfrac{-5}{8}-x=\dfrac{-3}{20}-\dfrac{-1}{6}\)
\(\dfrac{-5}{8}-x=\dfrac{1}{60}\)
\(\Rightarrow x=\dfrac{-5}{8}-\dfrac{1}{60}\)
\(\Rightarrow x=\dfrac{-77}{120}\)
d) \(\dfrac{3}{5}-x=\dfrac{1}{4}+\dfrac{7}{10}\)
\(\Rightarrow\dfrac{3}{5}-x=\dfrac{19}{20}\)
\(\Rightarrow x=\dfrac{3}{5}-\dfrac{19}{20}\)
\(\Rightarrow x=\dfrac{-7}{20}\)
e) \(\dfrac{-3}{7}-x=\dfrac{4}{5}+\dfrac{-2}{3}\)
\(\Rightarrow\dfrac{-3}{7}-x=\dfrac{2}{15}\)
\(\Rightarrow x=\dfrac{-3}{7}-\dfrac{2}{15}\)
\(\Rightarrow x=\dfrac{-59}{105}\)
g) \(\dfrac{-5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)
\(\Rightarrow\dfrac{-5}{6}-x=\dfrac{1}{4}\)
\(\Rightarrow x=\dfrac{-5}{6}-\dfrac{1}{4}\)
\(\Rightarrow x=\dfrac{-13}{12}\)
4)
Ta có x \(\in\)B(5) = {...; -5; 0; 5; 10; 15; ...}
và -17 < x < 15
=> x \(\in\){-15; -10; 5; 0; 5; 10}
Tổng các số nguyên x thoả mãn điều kiện cho trước là:
(-15) + (-10) + (-5) + 0 + 5 + 10 = (-15) + (-10 + 10) + (-5 + 5) + 0 = -15
a) \(\dfrac{1}{2}-\left(x+\dfrac{1}{3}\right)=\dfrac{5}{6}\)
\(\Rightarrow x+\dfrac{1}{3}=\dfrac{1}{2}-\dfrac{5}{6}\)
\(\Rightarrow x+\dfrac{1}{3}=\dfrac{-1}{3}\)
\(\Rightarrow x=\dfrac{-1}{3}-\dfrac{1}{3}\)
\(\Rightarrow x=\dfrac{-2}{3}\)
b)\(\dfrac{3}{4}-\left(x+\dfrac{1}{2}\right)=\dfrac{4}{5}\)
\(\Rightarrow x+\dfrac{1}{2}=\dfrac{3}{4}-\dfrac{4}{5}\)
\(\Rightarrow x+\dfrac{1}{2}=\dfrac{-1}{20}\)
\(\Rightarrow x=\dfrac{-1}{20}-\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{-11}{20}\)
c) \(\dfrac{3}{35}-\left(\dfrac{3}{5}+x\right)=\dfrac{2}{7}\)
\(\Rightarrow\dfrac{3}{5}+x=\dfrac{3}{35}-\dfrac{2}{7}\)
\(\Rightarrow\dfrac{3}{5}+x=\dfrac{-1}{5}\)
\(\Rightarrow x=\dfrac{-1}{5}-\dfrac{3}{5}\)
\(\Rightarrow x=\dfrac{-4}{5}\)
d)\(\dfrac{2}{3}.x=\dfrac{4}{27}\)
\(\Rightarrow x=\dfrac{4}{27}:\dfrac{2}{3}\)
\(\Rightarrow x=\dfrac{2}{9}\)
e) \(\dfrac{-3}{5}.x=\dfrac{21}{10}\)
\(\Rightarrow x=\dfrac{21}{10}:\dfrac{-3}{5}\)
\(\Rightarrow x=\dfrac{-7}{2}\)
C=(1x3+3x5+...+99x101)+(2x4+4x6+...+98x100)
đặt S=1x3+3x5+...+99x101
=>6S=6x(1x3+3x5+...+99x101)
=1x3x(5+1)+3x5x(7-1)+...+97x99x(101-95)+99x101x(103-97)
=1x3x5+1x3x1+3x5x7-1x3x5+....+97x99x101-95x97x99+99x101x103-97x99x101
=1x3x1+99x101x103
=>S=(3+99x101x103):6=171650
=>C=171650+(2x4+4x6+...+98x100)
đặt A=2x4+4x6+...+98x100
=>6A=6x(2x4+4x6+...+98x100)
=>6A=2x4x6+4x6x(8-2)+...+96x98x(100-94)+98x100x(102-96)
=2x4x6+4x6x8-2x4x6+...+96x98x100-94x96x98+98x100x102-96x98x100
=98x100x102
=>A=98x100x102:6=166600
=>C=166600+171650
=>C=338250
B=2x2+4x4+6x6+...+100x100
=2x(4-2)+4x(6-2)+6x(8-2)+...+100x(102-2)
=2x4-4+4x6-8+6x8-12+...+100x102-200
=(2x4+4x6+6x8+...+100x102)-(4+8+12+...+200)
đặt A=2x4+4x6+...+98x100+100x102
=>6A=6x(2x4+4x6+...+98x100+100x102)
=>6A=2x4x6+4x6x(8-2)+...+96x98x(100-94)+98x100x(102-96)+100x102x(104-98)
=2x4x6+4x6x8-2x4x6+...+96x98x100-94x96x98+98x100x102-96x98x100+100x102x104-98x100x102
=100x102x104
=>A=100x102x104:6=176800
=>B=176800-(4+8+12+...+200)
đặt S=4+8+12+..+200
Số số hạng của S là:
(200-4):4+1=50 số
S=(200+4)x50:2=5100
=>B=176800-5100
=>B=171700
a) \(x-\dfrac{3}{5}=\dfrac{4}{-10}\)
\(x=\dfrac{4}{-10}+\dfrac{3}{5}\)
\(x=\dfrac{-4}{10}+\dfrac{6}{10}\)
\(x=\dfrac{1}{5}\)
b) \(\dfrac{3}{x}-2=\dfrac{4}{x}+4\)
\(\dfrac{3}{x}-2+2=\dfrac{4}{x}+4+2\)
\(\dfrac{3}{x}=\dfrac{4}{x}+4\)
\(\dfrac{3}{x}=\dfrac{4x+4}{x}\)
\(3x=\left(4x+4\right)x\)
\(3x=5x\cdot x+4x\)
\(3x=x\left(5x+4\right)\)
\(3=5x+4\)
\(5x=-1\)
\(x=\dfrac{-1}{5}\)
Toán này lớp 6 á???