I x + 3/4 l - l x-7/3 l = 0
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QD
22 tháng 6 2017
d, \(\left(3x-2^4\right).7^3=2.7^4\)
\(\Rightarrow3x-2^4=2.7^4:7^3\)
\(\Rightarrow3x-16=2.7\\ \Rightarrow3x=14+16\\ \Rightarrow3x=30\Rightarrow x=10\)
Vậy.....
e, \(x-\left[42+\left(-28\right)\right]=-8\)
\(\Rightarrow x-14=-8\\ \Rightarrow x=6\)
Vậy.....
g, \(x-7=-5\)
\(\Rightarrow x=-5+7\Rightarrow x=2\)
Vậy.....
h, \(15-5\left(x+4\right)=-12-3\)
\(\Rightarrow15-5x-20=-15\)
\(\Rightarrow-5x=-15-15+20\)
\(\Rightarrow-5x=-10\Rightarrow x=2\)
Vậy.....
Chúc bạn học tốt!!!
AT
22 tháng 6 2017
d/ \(\left(3x-2^4\right)\cdot7^3=2\cdot7^4\)
\(\Rightarrow3x-16=\dfrac{2\cdot7^4}{7^3}=14\)
\(\Rightarrow3x=14+16=30\)
\(\Rightarrow x=\dfrac{30}{3}=10\)
e/ Đễ ==> tự lm thì tốt hơn nhé
g/ Đễ ==> tự lm thì tốt hơn nhé
h/ \(15-5\left(x+4\right)=-12-3\)
\(\Rightarrow15-5x-20=-15\)
\(\Rightarrow-5x=-15+20-15=-10\)
\(\Rightarrow x=\dfrac{-10}{-5}=2\)
i/ \(\left(7-x\right)-\left(25+7\right)=-25\)
\(\Rightarrow7-x-25-7=-25\)
\(\Rightarrow-x=-25-7+7+25\)
\(\Rightarrow-x=0\Rightarrow x=0\)
k/ \(\left|x+2\right|=0\Rightarrow x+2=0\Rightarrow x=-2\)
l/ \(\left|x-3\right|=7-\left(-2\right)\)
\(\Rightarrow\left|x-3\right|=9\)
\(\Rightarrow\left[{}\begin{matrix}x-3=9\\x-3=-9\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=12\\x=-6\end{matrix}\right.\)
m/ \(\left|x-5\right|=\left|-7\right|\Rightarrow\left|x-5\right|=7\)
\(\Rightarrow\left[{}\begin{matrix}x-5=7\\x-5=-7\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=12\\x=-2\end{matrix}\right.\)
11 tháng 7 2015
1) 2. I2x-3l = 1/2
|2x-3| =1/2:2
|2x-3| =1/4
=>2x-3 =1/4 hoặc 2x-3 =-1/4
2x =1/4+3 2x =-1/4+3
2x =13/4 2x =11/4
x =13/4:2 x =11/4:2
x =13/8 x =11/8
vậy x=13/8 hoặc 11/8
tich dung cho minh nhe
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
17 tháng 3 2024
a; |\(x+2\)| = 0
\(x+2=0\)
\(x\) = - 2
Vậy \(x\) = - 2
b; |\(x-5\)| = |-7|
| \(x-5\) | = 7
\(\left[{}\begin{matrix}x=7\\x=-7\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=5+7\\x=-7+5\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=12\\x=-2\end{matrix}\right.\)
Vậy \(x=12\)
\(x=-2\)
13 tháng 6 2017
a) Ta có: \(\left|x+\dfrac{19}{5}\right|\ge0\forall x\in Q\)
\(\left|y+\dfrac{2017}{2018}\right|\ge0\forall y\in Q\)
\(\left|z-2019\right|\ge0\forall x\in Q\)
\(\Rightarrow\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{2017}{2018}\right|+\left|z-2019\right|\ge0\forall x,y,z\in Q\)
Dấu \("="\) xảy ra khi \(\left\{{}\begin{matrix}\left|x+\dfrac{19}{5}\right|=0\\\left|y+\dfrac{2017}{2018}\right|=0\\\left|z-2019\right|=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{-19}{5}\\y=\dfrac{-2017}{2018}\\z=2019\end{matrix}\right.\).
b) Lại có:
\(\left|x-\dfrac{9}{5}\right|\ge0\forall x\in Q\)
\(\left|y+\dfrac{3}{4}\right|\ge0\forall y\in Q\)
\(\left|z+\dfrac{7}{2}\right|\ge0\forall z\in Q\)
\(\Rightarrow\left|x-\dfrac{9}{5}\right|+\left|y+\dfrac{3}{4}\right|+\left|z+\dfrac{7}{2}\right|\ge0\forall x,y,zQ\)
Mà theo đề bài:
\(\left|x-\dfrac{9}{5}\right|+\left|y+\dfrac{3}{4}\right|+\left|z+\dfrac{7}{2}\right|\le0\forall\)
\(\Rightarrow\left|x-\dfrac{9}{5}\right|+\left|y+\dfrac{3}{4}\right|+\left|z+\dfrac{7}{2}\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|x-\dfrac{9}{5}\right|=0\\\left|y+\dfrac{3}{4}\right|=0\\\left|z+\dfrac{7}{2}\right|=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{9}{5}\\y=\dfrac{-3}{4}\\z=\dfrac{-7}{2}\end{matrix}\right.\)
Vậy .....
PT
13 tháng 6 2017
a) \(\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{2017}{2018}\right|+\left|z-2019\right|=0\)
Ta có: \(\left|x+\dfrac{19}{5}\right|\ge0;\left|y+\dfrac{2017}{2018}\right|\ge0;\left|z-2019\right|\ge0\)
Để \(\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{2017}{2018}\right|+\left|z-2019\right|=0\) thì:
\(\left\{{}\begin{matrix}\left|x+\dfrac{19}{5}\right|=0\\\left|y+\dfrac{2017}{2018}\right|=0\\\left|z-2019\right|=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{-19}{5}\\y=\dfrac{-2017}{2018}\\z=2019\end{matrix}\right.\)
Vậy............................
b) Ta có: \(\left|x-\dfrac{9}{5}\right|\ge0;\left|y+\dfrac{3}{4}\right|\ge0;\left|z+\dfrac{7}{2}\right|\ge0\)
Mà \(\left|x-\dfrac{9}{5}\right|+\left|y+\dfrac{3}{4}\right|+\left|z+\dfrac{7}{2}\right|\le0\) thì:
\(\left|x-\dfrac{9}{5}\right|=\left|y+\dfrac{3}{4}\right|=\left|z+\dfrac{7}{2}\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{9}{5}\\y=\dfrac{-3}{4}\\z=\dfrac{-7}{2}\end{matrix}\right.\)
Vậy............................
NV
Nguyễn Việt Lâm
Giáo viên
9 tháng 3 2020
a/ \(log_2\left(2+a\right)=3\Rightarrow2+a=8\Rightarrow a=6\)
b/ Đặt \(\left(2+\sqrt{3}\right)^x=t>0\)
\(\Rightarrow t^2+t=6\Leftrightarrow t^2+t-6=0\Rightarrow\left[{}\begin{matrix}t=-3\left(l\right)\\t=2\end{matrix}\right.\)
\(\Rightarrow\left(2+\sqrt{3}\right)^x=2\Rightarrow x=log_{2+\sqrt{3}}2\)
c/ Đặt \(2^x=t>0\)
\(t^2-5t+4=0\Rightarrow\left[{}\begin{matrix}t=1\\t=4\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}2^x=1\\2^x=4\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
13 tháng 12 2017
bài2
a, x-15=-63-4
=>x-15=-67
=>x=-52
b, -x+3=11
=>x=-11+3
=>x=-8
c,\(|\)x+2\(|\)-4=7
=>\(|\)x+2\(|\)=11
=>\(\left\{{}\begin{matrix}x+2=11\\x+2=-11\end{matrix}\right.\)=>\(\left\{{}\begin{matrix}x=9\\x=13\end{matrix}\right.\)
bài3
ta có:\(\left|y\right|\)=8
=>\(\left[{}\begin{matrix}y=8\\y=-8\end{matrix}\right.\)
TH1 x=5,y=8
=>x-y=5-8=-3
y-x=8-5=3
TH2x=5 ,y=-8
x-y=5--8=13
y-x=-8-5=-13
13 tháng 12 2017
Baif:
a) x-15=-63-4
x-15=-67
x=-67+15
x=-52
b)-x+3=11
-x=11-3
-x=8
=> x=8
c)\(\left|x+2\right|-4=7\)
\(\left|x+2\right|\)=7+4=11
=> x+2=11 hoặc x+2=-11
x=11-2=9 hoặc x=-11-2=-13
Bài 3:
TH1: Nếu x=5 và y=8
thì x-y=5-8=-3
y-x=8-5=3
TH
: Nếu x=5 và y=-8
thì x-y=5-(-8)=13
y-x=(-8)-5=-13
NT
1
MH
6 tháng 1 2016
a. x.(3-x) > 0
+) x > 0; 3-x > 0
=> x > 0; x < 3
=> 0 < x < 3
=> x thuộc {1; 2}
+) x < 0; 3-x < 0
=> x < 0; x > 3
=> 3 < x < 0 ( vô lí, loại )
Vậy x thuộc {1; 2}.
b. (x-72)2=4
=> (x-72)2=22=(-2)2
+) x-72=2
=> x=2+72
=> x=51
+) x-72=-2
=> x=-2+72
=> x=47
Vậy x thuộc {47; 51}.
c. |-5| - x = 16
=> 5 - x = 16
=> x = 5 - 16
=> x = -11
Nhanh lên nhé