Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: =>|x-1/4|=3/4
=>x-1/4=3/4 hoặc x-1/4=-3/4
=>x=1 hoặc x=-1/2
b: \(\left|x+\dfrac{1}{2}\right|=\dfrac{1}{2}-\dfrac{9}{4}=\dfrac{2-9}{4}=-\dfrac{7}{4}\)(vô lý)
c: \(\Leftrightarrow\left[{}\begin{matrix}2x+5=1-x\\2x+5=x-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-4\\x=-6\end{matrix}\right.\Leftrightarrow x\in\left\{-\dfrac{4}{3};-6\right\}\)
e: =>|3/2-x|=0
=>3/2-x=0
hay x=3/2
Bài 1:
a) \(x^2-3=1\)
\(\Rightarrow x^2=1+3=4\)
\(\Rightarrow x=\pm2\)
b)\(2x^3+12=-4\)
\(\Rightarrow2x^3=-4-12=-16\)
\(\Rightarrow x^3=-8\)
\(\Rightarrow x=-2\)
c)\(\left(2x-3\right)^2=16\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=4\\2x-3=-4\end{matrix}\right.\Leftrightarrow}\left[{}\begin{matrix}x=\dfrac{7}{2}\\-\dfrac{1}{2}\end{matrix}\right.\)
a) \(x^2-3=1\Rightarrow x^2=4\Rightarrow x=\pm2\)
b) \(2x^3+12=-4\Rightarrow2x^3=-16\)
\(\Rightarrow x^3=-\dfrac{16}{2}=-8=-2^3\)
\(\Rightarrow x=-2\)
c) \(\left(2x-3\right)^2=16\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=4\\2x-3=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
d,h,i,k cững tương tự....
|2x-1|=1,5
TH(1)2x-1=1,5
2x =1,5+1
2x =2,5
x =2,5 :2
x =1,25
TH(2) 2x-1=-1,5
2x =-1,5+1
2x =-0,5
x =-0,5:2
x =-0,25
các câu khác cứ tương tự bạn nhé
b) \(7,5-\left|5-2x\right|=-4,5\)
\(\left|5-2x\right|=7,5+4,7\)
\(\left|5-2x\right|=12\)
th1 :\(5-2x=12\)
\(2x=5-12\)
\(2x=-7\)
\(x=-7:2\)
\(x=-3,5\)
th2: \(5-2x=-12\)
\(2x=5+12\)
\(2x=17\)
\(x=17:2\)
\(x=8,5\)
c) \(-3+\left|x\right|=-1\)
\(\left|x\right|=-1+3\)
\(\left|x\right|=2\)
th1: \(x=-2\)
th2 : \(x=2\)
d)\(\left|2\dfrac{1}{3}-x\right|=\dfrac{1}{6}\)
\(\left|\dfrac{7}{3}-x\right|=\dfrac{1}{6}\)
th1 :\(\dfrac{7}{3}-x=\dfrac{1}{6}\)
\(x=\dfrac{7}{3}-\dfrac{1}{2}\)
\(x=\dfrac{11}{6}\)
th2: \(\dfrac{7}{3}-x=\dfrac{-1}{6}\)
\(x=\dfrac{7}{3}+\dfrac{1}{6}\)
\(x=\dfrac{-5}{2}\)
e) \(\dfrac{5}{7}-\left|x+1\right|=\dfrac{1}{14}\)
\(\left|x+1\right|=\dfrac{5}{7}-\dfrac{1}{14}\)
\(\left|x+1\right|=\dfrac{9}{14}\)
th1 :\(x+1=\dfrac{9}{14}\)
\(x=\dfrac{9}{14}-1\)
\(x=\dfrac{-5}{14}\)
th2 : \(x+1=\dfrac{-9}{14}\)
\(x=\dfrac{-9}{14}-1\)
\(x=\dfrac{-5}{14}\)
h) \(5^x+5^{x+2}=650\)
\(\Leftrightarrow5^x+5^x.5^2=650\)
\(\Leftrightarrow5^x\left(1+25\right)=650\)
\(\Leftrightarrow5^x.26=650\)
\(\Leftrightarrow5^x=25\)
\(\Leftrightarrow x=2\)
haizzz,đăng ít thôi,chứ nhìn hoa mắt quá =.=
bây định làm j ở chỗ này vậy??? có j ib ns vs nhao chớ sao ns ở đây
\(a,\frac{15^3.\left(-5\right)^4}{\left(-3\right)^5.5^6}\)\(=\frac{3^3.5^3}{\left(-3\right)^5.5^2}\)\(=-\frac{5}{\left(3\right)^2}=-\frac{5}{9}\)
\(b,\frac{6^3.2.\left(-3\right)^2}{\left(-2\right)^9.3^7}\)\(=-\frac{6^3}{2^8.3^5}\)\(=-\frac{2^3.3^3}{2^8.3^5}\)\(=-\frac{1}{2^5.3^2}=-\frac{1}{288}\)
\(c,\frac{3^6.7^2-3^7.7}{3^7.21}\)\(=\frac{3^6.7\left(7-3\right)}{3^7.21}\)\(=\frac{3^6.7.4}{3^7.7.3}\)\(=\frac{4}{3.3}=\frac{4}{9}\)
\(a,\left(x-1,2\right)^2=4\)
\(\Rightarrow x-1,2=2\)
\(\Rightarrow x=3,2\)
\(b,\left(x+1\right)^3=-125\)
\(\Rightarrow\left(x+1\right)^3=\left(-5\right)^3\)
\(\Rightarrow x+1=-5\Rightarrow x=-6\)
\(c,\left(x-5\right)^3=2^6\)
\(\Rightarrow\left(x-5\right)^3=4^3\)
\(\Rightarrow x-5=4\Rightarrow x=9\)
\(d,\left(2x+1\right)^{x+1}=5^{x+1}\)
\(\Rightarrow2x+1=5\Rightarrow x=2\)
b: \(\left(\dfrac{2}{5}-\dfrac{7}{10}x\right):\dfrac{5}{3}=-\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{2}{5}-\dfrac{7}{10}x=\dfrac{-3}{4}\cdot\dfrac{5}{3}=\dfrac{-5}{4}\)
\(\Leftrightarrow x\cdot\dfrac{7}{10}=\dfrac{2}{5}+\dfrac{5}{4}=\dfrac{8+25}{20}=\dfrac{33}{20}\)
\(\Leftrightarrow x=\dfrac{33}{20}:\dfrac{7}{10}=\dfrac{33}{20}\cdot\dfrac{10}{7}=\dfrac{33}{14}\)
c: \(\dfrac{7}{16}:\left(\dfrac{1}{4}x+\dfrac{9}{2}\right)-\dfrac{11}{6}=0\)
\(\Leftrightarrow\dfrac{7}{16}:\left(\dfrac{1}{4}x+\dfrac{9}{2}\right)=\dfrac{11}{6}\)
\(\Leftrightarrow x\cdot\dfrac{1}{4}+\dfrac{9}{2}=\dfrac{11}{6}:\dfrac{7}{16}=\dfrac{88}{21}\)
\(\Leftrightarrow x\cdot\dfrac{1}{4}=\dfrac{88}{21}-\dfrac{9}{2}=-\dfrac{13}{42}\)
hay \(x=-\dfrac{26}{21}\)
a)
Ta thấy \(\left\{\begin{matrix} |x+\frac{19}{5}|\geq 0\\ |y+\frac{1890}{1975}|\geq 0\\ |z-2005|\geq 0\end{matrix}\right., \forall x,y,z\in\mathbb{Z}\)
\(|x+\frac{19}{5}|+|y+\frac{1890}{1975}|+|z-2005|\geq 0\)
Do đó, để \(|x+\frac{19}{5}|+|y+\frac{1890}{1975}|+|z-2005|=0\) thì :
\(\left\{\begin{matrix} |x+\frac{19}{5}|= 0\\ |y+\frac{1890}{1975}|= 0\\ |z-2005|=0\end{matrix}\right.\Rightarrow x=\frac{-19}{5}; y=\frac{-1890}{1975}; z=2005\)
b) Giống phần a, vì trị tuyệt đối của một số luôn không âm nên để tổng các trị tuyệt đối bằng $0$ thì:
\(\left\{\begin{matrix} |x+\frac{3}{4}|=0\\ |y-\frac{1}{5}|=0\\ |x+y+z|=0\end{matrix}\right.\Rightarrow \left\{\begin{matrix} x=-\frac{3}{4}\\ y=\frac{1}{5}\\ z=-(x+y)=\frac{11}{20}\end{matrix}\right.\)
c) \(\frac{16}{2^x}=1\Rightarrow 16=2^x\)
\(\Leftrightarrow 2^4=2^x\Rightarrow x=4\)
d) \((2x-1)^3=-27=(-3)^3\)
\(\Rightarrow 2x-1=-3\)
\(\Rightarrow 2x=-2\Rightarrow x=-1\)
e) \((x-2)^2=1=1^2=(-1)^2\)
\(\Rightarrow \left[\begin{matrix} x-2=1\\ x-2=-1\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=3\\ x=1\end{matrix}\right.\)
f) \((x+\frac{1}{2})^2=\frac{4}{25}=(\frac{2}{5})^2=(\frac{-2}{5})^2\)
\(\Rightarrow \left[\begin{matrix} x+\frac{1}{2}=\frac{2}{5}\\ x+\frac{1}{2}=-\frac{2}{5}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{-1}{10}\\ x=\frac{-9}{10}\end{matrix}\right.\)
g) \((x-1)^2=(x-1)^6\)
\(\Leftrightarrow (x-1)^6-(x-1)^2=0\)
\(\Leftrightarrow (x-1)^2[(x-1)^4-1]=0\)
\(\Rightarrow \left[\begin{matrix} (x-1)^2=0\\ (x-1)^4=1=(-1)^4=1^4\end{matrix}\right.\)
\(\Rightarrow \left[\begin{matrix} x=1\\ \left[\begin{matrix} x-1=-1\\ x-1=1\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow \left[\begin{matrix} x=1\\ \left[\begin{matrix} x=0\\ x=2\end{matrix}\right.\end{matrix}\right.\)
Vậy \(x=\left\{0;1;2\right\}\)
a: \(\Leftrightarrow3^x\cdot3+2x\cdot3^x-18x-27=0\)
\(\Leftrightarrow3^x\left(2x+3\right)-9\left(2x+3\right)=0\)
\(\Leftrightarrow\left(2x+3\right)\left(3^x-9\right)=0\)
=>x=2 hoặc x=-3/2
b: \(\Leftrightarrow\left|2x+5\right|\cdot\dfrac{1}{2}-\dfrac{5}{4}\cdot2\cdot\left|2x+5\right|+\dfrac{7}{3}\cdot4\cdot\left|2x+5\right|=\dfrac{1}{6}\)
\(\Leftrightarrow\left|2x+5\right|=\dfrac{1}{44}\)
=>2x+5=1/44 hoặc 2x+1=-1/44
=>x=-219/88 hoặc x=-221/88
a)
\(\left|\dfrac{1}{2}-\dfrac{1}{3}+x\right|=-\dfrac{1}{4}-y\\ \Rightarrow\left\{{}\begin{matrix}\dfrac{1}{2}-\dfrac{1}{3}+x=-\dfrac{1}{4}-y\\\dfrac{1}{2}-\dfrac{1}{3}+x=\dfrac{1}{4}+y\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x+y=-\dfrac{5}{12}\\x-y=\dfrac{1}{12}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{6}\\y=-\dfrac{1}{4}\end{matrix}\right.\)
b)\(\left|x-y\right|+\left|y+\dfrac{9}{25}\right|=0\)
ta thấy : \(\left|x-y\right|\ge0\\ \left|y+\dfrac{9}{25}\right|\ge0\)\(\Rightarrow\left|x-y\right|+\left|y+\dfrac{9}{25}\right|\ge0\)
đẳng thửc xảy ra khi : \(\left\{{}\begin{matrix}x-y=0\\y+\dfrac{9}{25}=0\end{matrix}\right.\Rightarrow x=y=-\dfrac{9}{25}\)
vậy \(\left(x;y\right)=\left(-\dfrac{9}{25};-\dfrac{9}{25}\right)\)
c) \(\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}=0\)
ta thấy \(\left(\dfrac{1}{2}x-5\right)^{20}\:và\:\left(y^2-\dfrac{1}{4}\right)^{10}\) là các lũy thừa có số mũ chẵn
\(\Rightarrow\:\)\(\left(\dfrac{1}{2}x-5\right)^{20}\ge0\\ \left(y^2-\dfrac{1}{4}\right)^{10}\ge0\)\(\Rightarrow\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}\ge0\)
đẳng thức xảy ra khi \(\left\{{}\begin{matrix}\dfrac{1}{2}x-5=0\\y^2-\dfrac{1}{4}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=10\\\left[{}\begin{matrix}y=-\dfrac{1}{2}\\y=\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\)
vậy cặp số x,y cần tìm là \(\left(10;\dfrac{1}{2}\right)\:hoặc\:\left(10;-\dfrac{1}{2}\right)\)
d)
\(\left|x\left(x^2-\dfrac{5}{4}\right)\right|=x\\ \Leftrightarrow x\left(x^2-\dfrac{5}{4}\right)=x\left(vì\:x\ge0\right)\\ \Leftrightarrow x\left(x^2-\dfrac{9}{4}\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x^2-\dfrac{9}{4}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{3}{2}\end{matrix}\right.\end{matrix}\right.\)
vậy x cần tìm là \(-\dfrac{3}{2};0;\dfrac{3}{2}\)
e)\(x^2+\left(y-\dfrac{1}{10}\right)^4=0\)
ta thấy: \(x^2\ge0;\left(y-\dfrac{1}{10}\right)^4\ge0\)
\(\Rightarrow x^2+\left(y-\dfrac{1}{10}\right)^4\ge0\)
đẳng thức xảy ra khi: \(\left\{{}\begin{matrix}x=0\\y=\dfrac{1}{10}\end{matrix}\right.\)
vậy cặp số cần tìm là \(0;\dfrac{1}{10}\)
mình làm lại câu b) nha
b) |x-3|=-4
th1: x-3=-4
x=3+(-4)
x=-1
th2: x-3=4
x=3+4
x=7
b) \(\left|x-3\right|=-4\)
t/h1:\(x-3=-4\)
\(x=3-\left(-4\right)\)
\(x=7\)
t/h2:\(x-3=4\)
\(x=3-4\)
\(x=-1\)