a) 8/7 -1/7 : ( x/3 -2) = -1
b) ( /x/ +1/3)2 = ( -1/2)3
Ai giúp mình với ạ
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(\dfrac{13}{20}+\dfrac{3}{5}+x=\dfrac{5}{6}\)
\(\Rightarrow\dfrac{5}{4}+x=\dfrac{5}{6}\)
\(\Rightarrow x=\dfrac{5}{6}-\dfrac{5}{4}\)
\(\Rightarrow x=\dfrac{-5}{12}\)
b) \(x+\dfrac{1}{3}=\dfrac{2}{5}-\dfrac{-1}{3}\)
\(\Rightarrow x+\dfrac{1}{3}=\dfrac{11}{15}\)
\(\Rightarrow x=\dfrac{11}{15}-\dfrac{1}{3}\)
\(\Rightarrow x=\dfrac{2}{5}\)
c)\(\dfrac{-5}{8}-x=\dfrac{-3}{20}-\dfrac{-1}{6}\)
\(\dfrac{-5}{8}-x=\dfrac{1}{60}\)
\(\Rightarrow x=\dfrac{-5}{8}-\dfrac{1}{60}\)
\(\Rightarrow x=\dfrac{-77}{120}\)
d) \(\dfrac{3}{5}-x=\dfrac{1}{4}+\dfrac{7}{10}\)
\(\Rightarrow\dfrac{3}{5}-x=\dfrac{19}{20}\)
\(\Rightarrow x=\dfrac{3}{5}-\dfrac{19}{20}\)
\(\Rightarrow x=\dfrac{-7}{20}\)
e) \(\dfrac{-3}{7}-x=\dfrac{4}{5}+\dfrac{-2}{3}\)
\(\Rightarrow\dfrac{-3}{7}-x=\dfrac{2}{15}\)
\(\Rightarrow x=\dfrac{-3}{7}-\dfrac{2}{15}\)
\(\Rightarrow x=\dfrac{-59}{105}\)
g) \(\dfrac{-5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)
\(\Rightarrow\dfrac{-5}{6}-x=\dfrac{1}{4}\)
\(\Rightarrow x=\dfrac{-5}{6}-\dfrac{1}{4}\)
\(\Rightarrow x=\dfrac{-13}{12}\)
\(a,ĐK:x\le\dfrac{5}{3}\\ PT\Leftrightarrow-3x+5=49\\ \Leftrightarrow x=-\dfrac{44}{3}\left(tm\right)\\ b,ĐK:x\ge-12\\ PT\Leftrightarrow\dfrac{1}{2}x+6=2\\ \Leftrightarrow\dfrac{1}{2}x=-4\\ \Leftrightarrow x=-8\left(tm\right)\\ c,ĐK:x\ge-\dfrac{1}{2}\\ PT\Leftrightarrow2x+1=13+4\sqrt{3}\\ \Leftrightarrow x=\dfrac{12+4\sqrt{3}}{2}=6+2\sqrt{3}\left(tm\right)\\ d,PT\Leftrightarrow\left|3x-1\right|=8\Leftrightarrow\left[{}\begin{matrix}3x-1=8\\1-3x=8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{7}{3}\end{matrix}\right.\)
A.
\(\Leftrightarrow\) 9x - 2x - 6 = 3x + 1
\(\Leftrightarrow\) 4x = 7
\(\Leftrightarrow\) x = \(\dfrac{7}{4}\)
B.
\(\Leftrightarrow\dfrac{5\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{4\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=\dfrac{x-13}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow\) 5x + 15 - 4x +12 = x - 13
\(\Leftrightarrow\) 0x = -40 ( phương trình vô nghiệm)
C.
\(\Leftrightarrow\) 7x + 8 \(\ge\) 3x -3
\(\Leftrightarrow\) 4x \(\ge\) - 11
\(\Leftrightarrow\)\(x\ge\dfrac{-11}{4}\)
1) \(A=36x^2+12x+1=\left(6x+1\right)^2\ge0\)
\(minA=0\Leftrightarrow x=-\dfrac{1}{6}\)
2) \(B=9x^2+6x+1=\left(3x+1\right)^2\ge0\)
\(minB=0\Leftrightarrow x=-\dfrac{1}{3}\)
4) \(D=x^2-4x+y^2-8y+6=\left(x-2\right)^2+\left(y-4\right)^2-14\ge-14\)
\(minD=-14\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=4\end{matrix}\right.\)
3) \(C=\left(x+1\right)\left(x-2\right)\left(x-3\right)\left(x-6\right)=\left(x^2-5x-6\right)\left(x^2-5x+6\right)=\left(x^2-5x\right)^2-36\ge-36\)
\(minC\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
5) \(E=\left(x-8\right)^2+\left(x+7\right)^2=2x^2-2x+113=2\left(x-\dfrac{1}{2}\right)^2+\dfrac{225}{2}\ge\dfrac{225}{2}\)
\(minE=\dfrac{225}{2}\Leftrightarrow x=\dfrac{1}{2}\)
\(a,\Leftrightarrow-\dfrac{1}{2}x=\dfrac{1}{4}\Leftrightarrow x=-\dfrac{1}{2}\\ b,\Leftrightarrow\dfrac{1}{6}:x=\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\Leftrightarrow x=\dfrac{1}{6}:\dfrac{5}{6}=\dfrac{1}{5}\\ c,\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=3\\x+\dfrac{1}{5}=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{14}{5}\\x=-\dfrac{16}{5}\end{matrix}\right.\)
\(d,\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\dfrac{22}{9}-\dfrac{7}{3}=\dfrac{1}{9}\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{3}\\x+\dfrac{1}{2}=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{6}\\x=-\dfrac{5}{6}\end{matrix}\right.\\ e,\Leftrightarrow2\left|x\right|=2-\dfrac{1}{2}=\dfrac{3}{2}\\ \Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{3}{2}\\2x=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=-\dfrac{3}{4}\end{matrix}\right.\)
\(f,\Leftrightarrow\left|x+\dfrac{1}{2}\right|=1+\dfrac{1}{6}=\dfrac{7}{6}\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{7}{6}\\x+\dfrac{1}{2}=-\dfrac{7}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)
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 \(x\in\left\{\dfrac{3}{4};-\dfrac{3}{4}\right\}\)
a: =>4/3x=7/9-4/9=1/3
=>x=1/4
b: =>5/2-x=9/14:(-4/7)=-9/8
=>x=5/2+9/8=29/8
c: =>3x+3/4=8/3
=>3x=23/12
hay x=23/36
d: =>-5/6-x=7/12-4/12=3/12=1/4
=>x=-5/6-1/4=-10/12-3/12=-13/12
a: \(5^{\left(x-2\right)\left(x+3\right)}=1\)
=>\(\left(x-2\right)\left(x+3\right)=0\)
=>\(\left[{}\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
c: \(\left|x^2+2x\right|+\left|y^2-9\right|=0\)
mà \(\left\{{}\begin{matrix}\left|x^2+2x\right|>=0\forall x\\\left|y^2-9\right|>=0\forall y\end{matrix}\right.\)
nên \(\left\{{}\begin{matrix}x^2+2x=0\\y^2-9=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\left(x+2\right)=0\\\left(y-3\right)\left(y+3\right)=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x\in\left\{0;-2\right\}\\y\in\left\{3;-3\right\}\end{matrix}\right.\)
d: \(2^x+2^{x+1}+2^{x+2}+2^{x+3}=120\)
=>\(2^x\left(1+2+2^2+2^3\right)=120\)
=>\(2^x\cdot15=120\)
=>\(2^x=8\)
=>x=3
e: \(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)
=>\(\left(x-7\right)^{x+11}-\left(x-7\right)^{x+1}=0\)
=>\(\left(x-7\right)^{x+1}\left[\left(x-7\right)^{10}-1\right]=0\)
=>\(\left[{}\begin{matrix}x-7=0\\x-7=1\\x-7=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=8\\x=6\end{matrix}\right.\)
A) \(x-\dfrac{2}{3}=\dfrac{4}{5}\\ x=\dfrac{4}{5}+\dfrac{2}{3}\)
\(x=\dfrac{22}{15}\)
b)\(\dfrac{7}{9}-x=\dfrac{1}{3}\\ x=\dfrac{7}{9}-\dfrac{1}{3}\\ x=\dfrac{4}{9}\)
C)\(x:\dfrac{2}{3}=\dfrac{9}{8}\\ x=\dfrac{9}{8}x\dfrac{2}{3}\\ x=\dfrac{3}{4}\)
a) 8/7 -1/7 : ( x/3 -2) = -1
1/7 : ( x/3 -2)=8/7-(-1)
1/7 : ( x/3 -2)=15/7
X/3-2=1/7:15/7
x/3-2=1/15
x/3=1/15+2
x/3=31/15
5x/15=31/15
5x=31
x=31/5
vậy......