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6 tháng 2 2018
a, => |5/3.x| = 1/6
=> 5/3.x = -1/6 hoặc 5/3.x = 1/6
=> x = -1/10 hoặc x = 1/10
Tk mk nha
23 tháng 9 2017
1) \(\left|x-\frac{3}{5}\right|< \frac{1}{3}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{3}{5}< \frac{1}{3}\\x-\frac{3}{5}< -\frac{1}{3}\end{cases}}\Rightarrow\orbr{\begin{cases}x< \frac{1}{3}+\frac{3}{5}\\x< \frac{-1}{3}+\frac{3}{5}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x< \frac{5}{15}+\frac{9}{15}\\x< \frac{-5}{15}+\frac{9}{15}\end{cases}}\Rightarrow\orbr{\begin{cases}x< \frac{14}{15}\\x< \frac{4}{15}\end{cases}}\)
vay \(\orbr{\begin{cases}x< \frac{14}{15}\\x< \frac{4}{15}\end{cases}}\)
2) \(\left|x+\frac{11}{2}\right|>\left|-5,5\right|\)
\(\left|x+\frac{11}{2}\right|>5,5\)
\(\Rightarrow\orbr{\begin{cases}x+\frac{11}{2}>\frac{11}{2}\\x+\frac{11}{2}>-\frac{11}{2}\end{cases}}\Rightarrow\orbr{\begin{cases}x>\frac{11}{2}-\frac{11}{2}\\x>\frac{-11}{2}-\frac{11}{2}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x>0\\x>-11\end{cases}}\)
vay \(\orbr{\begin{cases}x>0\\x>-11\end{cases}}\)
3) \(\frac{2}{5}< \left|x-\frac{7}{5}\right|< \frac{3}{5}\)
\(\Rightarrow\left|x-\frac{7}{5}\right|>\frac{2}{5}\) va \(\left|x-\frac{7}{5}\right|< \frac{3}{5}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{7}{5}>\frac{2}{5}\\x-\frac{7}{5}>\frac{-2}{5}\end{cases}}\Rightarrow\orbr{\begin{cases}x>\frac{2}{5}+\frac{7}{5}\\x>\frac{-2}{5}+\frac{7}{5}\end{cases}}\)va \(\orbr{\begin{cases}x-\frac{7}{5}< \frac{3}{5}\\x-\frac{7}{5}< \frac{-3}{5}\end{cases}}\Rightarrow\orbr{\begin{cases}x< \frac{3}{5}+\frac{7}{5}\\x< \frac{-3}{5}+\frac{7}{5}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x>\frac{9}{5}\\x>1\end{cases}}\)va \(\orbr{\begin{cases}x< 2\\x< \frac{4}{5}\end{cases}}\)
vay ....
30 tháng 7 2018
\(\left(x+\frac{1}{2}\right)\left(x-\frac{3}{4}\right)=0\)
\(\Rightarrow\hept{\begin{cases}x+\frac{1}{2}=0\\x-\frac{3}{4}=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=-\frac{1}{2}\\x=\frac{3}{4}\end{cases}}\)
DH
25 tháng 6 2017
b, \(\dfrac{3}{\left(x+2\right)\left(x+5\right)}+\dfrac{5}{\left(x+5\right)\left(x+10\right)}+\dfrac{7}{\left(x+10\right)\left(x+17\right)}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow\dfrac{1}{x+2}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+10}+\dfrac{1}{x+10}-\dfrac{1}{x+17}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow\dfrac{1}{x+2}-\dfrac{1}{x+17}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow\dfrac{x+17-x+2}{\left(x+2\right)\left(x+17\right)}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow x=19\)
Chúc bạn học tốt!!!
25 tháng 6 2017
a, \(\dfrac{x+1}{5}+\dfrac{x+3}{4}=\dfrac{x+5}{3}+\dfrac{x+7}{2}\)
\(\Rightarrow\dfrac{x+1}{5}+2+\dfrac{x+3}{4}+2=\dfrac{x+5}{3}+2+\dfrac{x+7}{2}+2\)
\(\Rightarrow\dfrac{x+11}{5}+\dfrac{x+11}{4}-\dfrac{x+11}{3}-\dfrac{x+11}{2}=0\)
\(\Rightarrow\left(x+11\right)\left(\dfrac{1}{5}+\dfrac{1}{4}-\dfrac{1}{3}-\dfrac{1}{2}\right)=0\)
\(\Rightarrow x+11=0\Rightarrow x=-11\)
Vậy x = -11
b, \(\dfrac{3}{\left(x+2\right)\left(x+5\right)}+\dfrac{5}{\left(x+5\right)\left(x+10\right)}+\dfrac{7}{\left(x+10\right)\left(x+17\right)}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow\dfrac{1}{x+2}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+10}+\dfrac{1}{x+10}-\dfrac{1}{x+17}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow\dfrac{1}{x+2}-\dfrac{1}{x+17}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow\dfrac{x+17-x-2}{\left(x+2\right)\left(x+17\right)}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow\dfrac{15}{\left(x+2\right)\left(x+17\right)}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow x=15\)
Vậy x = 15
29 tháng 6 2019
câu a : Bạn lập bảng rồi tìm x,,y nhé
câu b :
\(x-\frac{3}{y}=\frac{x}{y}\)
\(\Leftrightarrow x=\frac{x}{y}+\frac{3}{y}\)
\(\Leftrightarrow x=\frac{3+x}{y}\)
\(\Leftrightarrow3+x=xy\)
\(\Leftrightarrow xy-x=3\)
\(\Leftrightarrow x.\left(y-1\right)=3\)
Lập bảng tìm x,y
NN
5 tháng 4 2020
1. \(A=x^{15}+3x^{14}+5=x^{14}\left(x+3\right)+5\)
Thay \(x+3=0\)vào đa thức ta được:\(A=x^{14}.0+5=5\)
2. \(B=\left(x^{2007}+3x^{2006}+1\right)^{2007}=\left[x^{2006}\left(x+3\right)+1\right]^{2007}\)
Thay \(x=-3\)vào đa thức ta được: \(B=\left[x^{2006}\left(-3+3\right)+1\right]^{2017}=\left(x^{2006}.0+1\right)^{2017}=1^{2017}=1\)
3. \(C=21x^4+12x^3-3x^2+24x+15=3x\left(7x^3+4x^2-x+8\right)+15\)
Thay \(7x^3+4x^2-x+8=0\)vào đa thức ta được: \(C=3x.0+15=15\)
4. \(D=-16x^5-28x^4+16x^3-20x^2+32x+2007\)
\(=4x\left(-4x^4-7x^3+4x^2-5x+8\right)+2007\)
Thay \(-4x^4-7x^3+4x^2-5x+8=0\)vào đa thức ta được: \(D=4x.0+2007=2007\)
5 tháng 4 2020
1. \(A=x^{15}+3x^{14}+5\)
\(A=x^{14}\left(x+3\right)+5\)
\(A=x^{14}+5\)
2. \(B=\left(x^{2007}+3x^{2006}+1\right)^{2007}\)
\(B=\left[x^{2006}\left(x+3\right)+1\right]^{2007}\)
\(B=\left[x^{2006}.\left(-3+3\right)+1\right]^{2007}\)
\(B=1^{2007}=1\)
3. \(C=21x^4+12x^3-3x^2+24x+15\)
\(C=3x\left(7x^2+4x^2-x+8+5\right)\)
\(C=3x\left(0+5\right)\)
\(C=15x\)
4. \(D=-16x^5-28x^4+16x^3-20x^2+32+2007\)
\(D=4x\left(-4x^4-7x^3+4x^2-5x+8\right)+2007\)
\(D=4x.0+2007\)
\(D=2007\)
9 tháng 7 2019
\(K=|x-1|+|x-2|+|x-3|\)
\(=\left(|x-1|+|x-3|\right)+|x-2|\)
\(=\left(|x-1|+|3-x|\right)+|x-2|\)
Đặt \(A=|x-1|+|3-x|\ge|x-1+3-x|\)
Hay \(A\ge2\left(1\right)\)
Dấu "= " xảy ra \(\Leftrightarrow\left(x-1\right)\left(3-x\right)\ge0\)
\(\Leftrightarrow\hept{\begin{cases}x-1\ge0\\3-x\ge0\end{cases}}\)hoặc \(\hept{\begin{cases}x-1< 0\\3-x< 0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\ge1\\x\le3\end{cases}}\)hoặc \(\hept{\begin{cases}x< 1\\x>3\end{cases}\left(loai\right)}\)
\(\Leftrightarrow1\le x\le3\)
Đặt \(B=|x-2|\)
Ta có: \(|x-2|\ge0;\forall x\)
Hay \(B\ge0;\forall x\left(2\right)\)
Dấu "=" xảy ra \(\Leftrightarrow|x-2|=0\)
\(\Leftrightarrow x=2\)
Từ \(\left(1\right);\left(2\right)\Rightarrow A+B\ge2+0\)
Hay \(K\ge2\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}1\le x\le3\\x=2\end{cases}\Leftrightarrow}x=2\)
Vậy MIN K=2 \(\Leftrightarrow x=2\)
31 tháng 7 2016
a) Lm r nkoa!
b) \(\left(0,25x\right):3=\frac{5}{6}:0,125\)
\(=\left(0,25x\right):3=\frac{20}{3}\)
\(\Rightarrow0,25x=\frac{20}{3}\cdot3=20\)
\(\Rightarrow x=20:0,25=80\)
\(\Rightarrow x=80\)
c) \(0,01:2,5=\left(0,75x\right):0,75\)
\(=\frac{1}{250}=\left(0,75x\right):0,75\)
\(\Rightarrow0,75x=\frac{1}{250}\cdot0,75=\frac{3}{1000}\)
\(\Rightarrow x=\frac{3}{1000}:0,75=\frac{1}{250}\)
\(\Rightarrow x=\frac{1}{250}\)
d) \(1\frac{1}{3}:0,8=\frac{2}{3}:\left(0,1x\right)\)
\(=\frac{5}{3}=\frac{2}{3}:\left(0,1x\right)\)
\(\Rightarrow0,1x=\frac{2}{3}:\frac{5}{3}=\frac{2}{5}\)
\(\Rightarrow x=\frac{2}{5}:0,1=4\)
\(\Rightarrow4\)
\(x^2\ge0\forall x\in R\Rightarrow|3+x^2|\ge3\forall x\in R.\)
\(\Rightarrow K\ge3+27,7=30,7.\)
Dấu "=" xảy ra <=> x=0