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a) x=-213:(1+2+3+4+...+100)<=>x=-213/100
b) x-x=-1/3-2/4 <=> 0= -5/6 (vô lý )
c) x=-0,8119408369
d) x= 0.0258907758
c: ĐKXĐ: x<>-1
Để C là số nguyên thì \(2x-7⋮x+1\)
=>\(2x+2-9⋮x+1\)
=>\(x+1\in\left\{1;-1;3;-3;9;-9\right\}\)
=>\(x\in\left\{0;-2;2;-4;8;-10\right\}\)
d: ĐKXĐ: x<>-3
Để D là số nguyên thì \(5x+9⋮x+3\)
=>\(5x+15-6⋮x+3\)
=>\(x+3\inƯ\left(-6\right)\)
=>\(x+3\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)
=>\(x\in\left\{-2;-4;-1;-5;0;-6;3;-9\right\}\)
\(\left(x-7\right)^{10}-\left(x-7\right)^{x+11}=0\)\(\Leftrightarrow\left(x-7\right)^{10}\left[1-\left(x-7\right)^{x+1}\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x-7\right)^{10}=0\\1-\left(x-7\right)^{x+1}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=7\\\left(x-7\right)^{x+1}=1\end{cases}}\)
Xét \(\left(x-7\right)^{x+1}=1\)ta có:
TH1: \(x+1=0\)và \(x-7\inℤ\)\(\Rightarrow x=-1\left(tm\right)\)
TH2: \(x-7=-1\)và \(x+1\)là số dương chẵn \(\Rightarrow x=6\left(tm\right)\)
TH3: \(x-7=1\)và \(x+1\inℕ^∗\) \(\Rightarrow x=8\left(tm\right)\)
Vậy \(x\in\left\{-1;6;7;8\right\}\)
Bạn có thể tự giải mấy bài cơ bản được không? Động não 1 chút nhé? bài nào không biết thật sự thì viết dưới bl t giải cho.Chứ dài z t không cân nổi:v
a.x+30/100x=-1,1
13/10x=-1,1
x=-11/13
b. (x-1/2):1/3 +5/7=9/5/7
(x-1/2):1/3=9
x-1/2=3
x=7/2
c. -5/6-x=7/12-1/3
x=-5/6-7/12-1/3
x=-7/4
d. 3(x+3)=-15
x+3=-5
x=-8
e. (4,5-2x)(-11/7)=11/14
4,5-2x=11/14:-11/7
4,5-2x=-1/2
2x=4,5+1/2
2x=5
x=5/2
`#3107.101107`
\(\dfrac{5}{7}\times\dfrac{6}{11}+\dfrac{5}{11}\times\dfrac{1}{7}-\dfrac{5}{7}\times\dfrac{14}{11}\\ =\dfrac{5}{7}\times\dfrac{6}{11}+\dfrac{5}{7}\times\dfrac{1}{11}-\dfrac{5}{7}\times\dfrac{14}{11}\\ =\dfrac{5}{7}\times\left(\dfrac{6}{11}+\dfrac{1}{11}-\dfrac{14}{11}\right)\\ =\dfrac{5}{7}\times\left(-\dfrac{7}{11}\right)\\ =-\dfrac{5}{11}\)
\(a,\Leftrightarrow-\dfrac{1}{2}x=\dfrac{1}{4}\Leftrightarrow x=-\dfrac{1}{2}\\ b,\Leftrightarrow\dfrac{1}{6}:x=\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\Leftrightarrow x=\dfrac{1}{6}:\dfrac{5}{6}=\dfrac{1}{5}\\ c,\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=3\\x+\dfrac{1}{5}=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{14}{5}\\x=-\dfrac{16}{5}\end{matrix}\right.\)
\(d,\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\dfrac{22}{9}-\dfrac{7}{3}=\dfrac{1}{9}\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{3}\\x+\dfrac{1}{2}=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{6}\\x=-\dfrac{5}{6}\end{matrix}\right.\\ e,\Leftrightarrow2\left|x\right|=2-\dfrac{1}{2}=\dfrac{3}{2}\\ \Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{3}{2}\\2x=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=-\dfrac{3}{4}\end{matrix}\right.\)
\(f,\Leftrightarrow\left|x+\dfrac{1}{2}\right|=1+\dfrac{1}{6}=\dfrac{7}{6}\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{7}{6}\\x+\dfrac{1}{2}=-\dfrac{7}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)
e: ta có: \(2\left|x\right|+\dfrac{1}{2}=2\)
\(\Leftrightarrow2\left|x\right|=\dfrac{3}{2}\)
\(\Leftrightarrow\left|x\right|=\dfrac{3}{4}\)
hay \(x\in\left\{\dfrac{3}{4};-\dfrac{3}{4}\right\}\)
Tìm x . biết :
\(a,\frac{2}{5}:\left(-x-\frac{1}{2}\right)=\frac{4}{5}\)
\(\Rightarrow-x-\frac{1}{2}=\frac{2}{5}:\frac{4}{5}\)
\(\Rightarrow-x-\frac{1}{2}=\frac{2}{5}.\frac{5}{4}\)
\(\Rightarrow-x-\frac{1}{2}=\frac{1}{2}\)
\(\Rightarrow-x=\frac{1}{2}+\frac{1}{2}\)
\(\Rightarrow-x=1\)
\(\Rightarrow x=-1\)
Vậy \(x=-1\)
a. \(\frac{2}{5}.\left(-x-\frac{1}{2}\right)=\frac{4}{5}\)
\(\Rightarrow-x-\frac{1}{2}=\frac{2}{5}:\frac{4}{5}\)
\(\Rightarrow-x-\frac{1}{2}=\frac{2}{5}.\frac{5}{4}\)
\(\Rightarrow-x-\frac{1}{2}=\frac{1}{2}\)
\(\Rightarrow-x=\frac{1}{2}+\frac{1}{2}\)
\(\Rightarrow-x=1\)
\(\Rightarrow x=-1\)
Đề bài yêu cầu gì?
GTNN của D