1. tim x biet:
a, 1+2+3+4+......+ x=1999
b, 1+4+7+.......+ x=103
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|x+1|+|3x-1|+|x-1|=3
=} vs cả trong dấu giá trị tuyệt đối >0 thì=}
x+1+3x-1+x-1=3{=}5x=4{=}x=4/5
=}vs cả trong giá trị tuyệt đối <0 thì=}
x+1+3x-1+x-1=-3{=}5x=-4{=}x=-4/5
b: \(\left(2x+1\right)^2=25\)
=>\(\left[{}\begin{matrix}2x+1=5\\2x+1=-5\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}2x=4\\2x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
c: \(\left(1-3x\right)^3=64\)
=>\(\left(1-3x\right)^3=4^3\)
=>1-3x=4
=>3x=1-4=-3
=>x=-3/3=-1
d: \(\left(4-x\right)^3=-27\)
=>\(\left(4-x\right)^3=\left(-3\right)^3\)
=>4-x=-3
=>x=4+3=7
e: \(x^2-5x=0\)
=>\(x\left(x-5\right)=0\)
=>\(\left[{}\begin{matrix}x=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
a: \(\Leftrightarrow x^2-x+4x-4+5⋮x-1\)
\(\Leftrightarrow x-1\in\left\{1;-1;5;-5\right\}\)
hay \(x\in\left\{2;0;6;-4\right\}\)
c: \(\Leftrightarrow4x-5=13k\left(k\in Z\right)\)
=>4k=13k+5
hay \(x=\dfrac{13k+5}{4}\)
1)\(x=\left(\dfrac{1}{5}+\dfrac{2}{3}\right):\dfrac{4}{7}=\left(\dfrac{3}{15}+\dfrac{10}{15}\right):\dfrac{4}{7}=\dfrac{13}{15}:\dfrac{4}{7}=\dfrac{91}{60}\)
2)\(x=\left(-\dfrac{3}{10}-\dfrac{3}{4}\right):\dfrac{1}{2}=\left(-\dfrac{6}{20}-\dfrac{15}{20}\right):\dfrac{1}{2}=-\dfrac{21}{20}:\dfrac{1}{2}=-\dfrac{21}{10}\)
3)\(x=-\dfrac{24}{36}-\dfrac{7}{15}=-\dfrac{2}{3}-\dfrac{7}{15}=\dfrac{-10}{15}-\dfrac{7}{15}=-\dfrac{17}{15}\)
1) \(\dfrac{4}{7}x=\dfrac{1}{5}+\dfrac{2}{3}\)
\(\dfrac{4}{7}x=\dfrac{13}{15}\)
\(x=\dfrac{91}{60}\)
2) \(\dfrac{1}{2}x=-\dfrac{3}{10}-\dfrac{3}{4}\)
\(\dfrac{1}{2}x=-\dfrac{21}{20}\)
\(x=-\dfrac{21}{10}\)
3) \(x=-\dfrac{24}{36}-\dfrac{7}{15}\)
\(x=\dfrac{-2}{3}+\dfrac{7}{15}\)
\(x=-\dfrac{17}{15}\)
a) \(\frac{3}{2}x-\frac{2}{5}=\frac{1}{3}x-\frac{1}{4}\)
\(\Rightarrow\frac{3}{2}x-\frac{1}{3}x=-\frac{1}{4}+\frac{2}{5}\)
\(\frac{7}{6}x=\frac{3}{20}\Rightarrow x=\frac{9}{70}\)
b) \(-5^{\frac{1}{2}x+1}=\frac{3}{4}-\frac{7}{6}\)
\(-5^{\frac{1}{2}x}.\left(-5\right)=-\frac{5}{12}\)
\(-5^{\frac{1}{2}x}=\frac{1}{12}\)
mà -51/2x mang giá trị âm
1/12 có giá trị dương
=> không tìm được x
c) \(\frac{2x-2}{3}=\frac{7x+3}{2-1}\)
\(\frac{2x-2}{3}=7x+3=\frac{21x+9}{3}\)
=> 2x - 2 = 21x + 9
=> 2x - 21x = 9 + 2
-19x = 11
x = -11/19
phần d bn lm như phần a nha
43553/5
35535