Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a. \(\dfrac{1}{2}x+\dfrac{3}{5}x=\dfrac{-33}{25}\)
\(\Rightarrow\dfrac{11}{10}x=\dfrac{-33}{25}\)
\(\Rightarrow x=\dfrac{-33}{25}:\dfrac{11}{10}=\dfrac{-6}{5}\)
Vậy.........
b. \(\left(\dfrac{2}{3}x-\dfrac{4}{9}\right)\left(\dfrac{1}{2}+\dfrac{-3}{7}:x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{2}{3}x-\dfrac{4}{9}=0\\\dfrac{1}{2}+\dfrac{-3}{7}:x=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\dfrac{2}{3}x=\dfrac{4}{9}\\\dfrac{-3}{7}:x=\dfrac{-1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=\dfrac{6}{7}\end{matrix}\right.\)
Vậy................
a: \(\dfrac{31-2x}{x+23}=\dfrac{9}{4}\)
=>121-8x=9x+207
=>-17x=86
hay x=-86/17
b: \(\dfrac{\left|2x-1\right|}{\dfrac{1}{2}}=\dfrac{18}{5}\)
=>|2x-1|=9/5
=>2x-1=9/5 hoặc 2x-1=-9/5
=>2x=14/5 hoặc 2x=-4/5
=>x=7/5 hoặc x=-2/5
Bài 4:
a) \(\dfrac{4}{3}+\left(1,25-x\right)=2,25\)
\(1,25-x=2,25-\dfrac{4}{3}=\dfrac{9}{4}-\dfrac{4}{3}\)
\(1,25-x=\dfrac{11}{12}\)
\(x=1,25-\dfrac{11}{12}=\dfrac{5}{4}-\dfrac{11}{12}\)
\(x=\dfrac{1}{3}\)
b) \(\dfrac{17}{6}-\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)
\(x-\dfrac{7}{6}=\dfrac{17}{6}-\dfrac{7}{4}=\dfrac{34}{12}-\dfrac{21}{12}\)
\(x-\dfrac{7}{6}=\dfrac{13}{12}\)
\(x=\dfrac{13}{12}+\dfrac{7}{6}=\dfrac{13}{12}+\dfrac{14}{12}\)
\(x=\dfrac{27}{12}=\dfrac{9}{4}\)
c) \(4-\left(2x+1\right)=3-\dfrac{1}{3}=\dfrac{9}{3}-\dfrac{1}{3}\)
\(4-\left(2x+1\right)=\dfrac{8}{3}\)
\(2x+1=\dfrac{8}{3}+4=\dfrac{8}{3}+\dfrac{12}{3}\)
\(2x+1=\dfrac{20}{3}\)
\(2x=\dfrac{20}{3}-1=\dfrac{20}{3}-\dfrac{3}{3}\)
\(2x=\dfrac{17}{3}\)
\(x=\dfrac{17}{3}.\dfrac{1}{2}=\dfrac{17}{6}\)
Bài 15:
a) \(\left(\dfrac{-2}{3}\right)^9:x=\dfrac{-2}{3}\)
\(x=\left(\dfrac{-2}{3}\right)^9:\dfrac{-2}{3}=\left(\dfrac{-2}{3}\right)^{9-1}\)
\(=>x=\left(\dfrac{-2}{3}\right)^8\)
b) \(x:\left(\dfrac{4}{9}\right)^5=\left(\dfrac{4}{9}\right)^4\)
\(x=\left(\dfrac{4}{9}\right)^4.\left(\dfrac{4}{9}\right)^5=\left(\dfrac{4}{9}\right)^{4+5}\)
\(=>x=\left(\dfrac{4}{9}\right)^9\)
c) \(\left(x+4\right)^3=-125\)
\(\left(x+4\right)^3=\left(-5\right)^3\)
\(=>x+4=-5\)
\(x=-5-4\)
\(=>x=-9\)
d) \(\left(10-5x\right)^3=64\)
\(\left(10-5x\right)^3=4^3\)
\(=>10-5x=4\)
\(5x=10-4\)
\(5x=6\)
\(=>x=\dfrac{6}{5}\)
e) \(\left(4x+5\right)^2=81\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(4x+5\right)^2=\left(-9\right)^2\\\left(4x+5\right)^2=9^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+5=-9\\4x+5=9\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=-14\\4x=4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-14}{4}\\x=1\end{matrix}\right.\)
Bài 16:
a) \(4-1\dfrac{2}{5}-\dfrac{8}{3}\)
\(=4-\dfrac{7}{5}-\dfrac{8}{3}\)
\(=\dfrac{60-21-40}{15}=\dfrac{-1}{15}\)
b) \(-0,6-\dfrac{-4}{9}-\dfrac{16}{15}\)
\(=\dfrac{-3}{5}+\dfrac{4}{9}-\dfrac{16}{15}\)
\(=\dfrac{\left(-27\right)+20-48}{45}=\dfrac{-55}{45}=\dfrac{-11}{9}\)
c) \(-\dfrac{15}{4}.\left(\dfrac{-7}{15}\right).\left(-2\dfrac{2}{5}\right)\)
\(=\dfrac{7}{4}.\dfrac{-12}{5}\)
\(=\dfrac{-21}{5}\)
\(#Wendy.Dang\)
\(\dfrac{5}{6}x-\dfrac{3}{4}=\dfrac{-1}{4}+\dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{5}{6}x=\dfrac{7}{6}\)
\(\Rightarrow x=\dfrac{7}{5}\)
b) \(-1\dfrac{1}{2}-\dfrac{2}{3}x=\dfrac{5}{6}-\left(\dfrac{-2}{5}\right)\)
\(\Leftrightarrow\dfrac{2}{3}x=-\dfrac{41}{15}\)
\(\Rightarrow x=-\dfrac{41}{10}\)
c) \(\left(\dfrac{4}{5}:x+1,5\right):\dfrac{2}{3}=-1,5\)
\(\Leftrightarrow\dfrac{8+15x}{10x}.\dfrac{3}{2}=\dfrac{-3}{2}\)
\(\Leftrightarrow\dfrac{24+45x}{20x}=\dfrac{-3}{2}\)
\(\Leftrightarrow-60x=48+90x\)
\(\Rightarrow x=-0,32\)
d) \(\dfrac{4}{3}x-\dfrac{2}{3}=\dfrac{1}{4}-x\)
\(\Leftrightarrow\dfrac{4x-2}{3}=\dfrac{1-4x}{4}\)
\(\Rightarrow16x-8=3-12x\)
\(\Rightarrow x=\dfrac{11}{28}\)
a, Ta có: \(\left(x-\dfrac{1}{2}\right)^2\ge0\)
\(\Leftrightarrow A=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
Dấu " = " khi \(\left(x-\dfrac{1}{2}\right)^2=0\Leftrightarrow x=\dfrac{1}{2}\)
Vậy \(MIN_A=\dfrac{3}{4}\) khi \(x=\dfrac{1}{2}\)
b, Để B lớn nhất thì \(\left(x-\dfrac{2}{3}\right)^2+9\) nhỏ nhất
Ta có: \(\left(x-\dfrac{2}{3}\right)^2+9\ge9\)
\(\Leftrightarrow B=\dfrac{4}{\left(x-\dfrac{2}{3}\right)^2+9}\le\dfrac{4}{9}\)
Dấu " = " khi \(\left(x-\dfrac{2}{3}\right)^2=0\Leftrightarrow x=\dfrac{2}{3}\)
Vậy \(MAX_B=\dfrac{4}{9}\) khi \(x=\dfrac{2}{3}\)
1: \(\left(\dfrac{1}{16}\right)^x=\left(\dfrac{1}{8}\right)^6\)
\(\Leftrightarrow\left(\dfrac{1}{2}\right)^{4x}=\left(\dfrac{1}{2}\right)^{18}\)
=>4x=18
hay x=9/2
2: \(\left(\dfrac{1}{16}\right)^x=\left(\dfrac{1}{8}\right)^{36}\)
\(\Leftrightarrow\left(\dfrac{1}{2}\right)^{4x}=\left(\dfrac{1}{2}\right)^{108}\)
=>4x=108
hay x=27
3: \(\left(\dfrac{1}{81}\right)^x=\left(\dfrac{1}{27}\right)^4\)
\(\Leftrightarrow\left(\dfrac{1}{3}\right)^{4x}=\left(\dfrac{1}{3}\right)^{12}\)
=>4x=12
hay x=3
a: \(x=\left(-\dfrac{2}{3}\right)^5:\left(-\dfrac{2}{3}\right)^2=\left(-\dfrac{2}{3}\right)^3=-\dfrac{8}{27}\)
b: =>x-1/2=1/3
=>x=5/6
c: =>2/3x-1=0 hoặc 3/4x+1/2=0
=>x=3/2 hoặc x=-1/2:3/4=-1/2*4/3=-4/6=-2/3
d =>4/9:x=10/3:9/4=10/3*4/9=40/27
=>x=4/9:40/27=4/9*27/40=108/360=3/10
1.a)\(2.x-\dfrac{5}{4}=\dfrac{20}{15}\)
\(\Leftrightarrow2.x=\dfrac{20}{15}+\dfrac{5}{4}=\dfrac{4}{3}+\dfrac{5}{4}=\dfrac{16+15}{12}=\dfrac{31}{12}\)
\(\Leftrightarrow x=\dfrac{31}{12}:2=\dfrac{31}{12}.\dfrac{1}{2}=\dfrac{31}{24}\)
b)\(\left(x+\dfrac{1}{3}\right)^3=\left(-\dfrac{1}{8}\right)\)
\(\Leftrightarrow\left(x+\dfrac{1}{3}\right)^3=\left(-\dfrac{1}{2}\right)^3\)
\(\Leftrightarrow x+\dfrac{1}{3}=-\dfrac{1}{2}\)
\(\Leftrightarrow x=-\dfrac{1}{2}-\dfrac{1}{3}=-\dfrac{5}{6}\)
2.Theo đề bài, ta có: \(\dfrac{a}{2}=\dfrac{b}{3}\) và \(a+b=-15\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{a+b}{2+3}=\dfrac{-15}{5}=-3\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{2}=-3\Rightarrow a=-6\\\dfrac{b}{3}=-3\Rightarrow b=-9\end{matrix}\right.\)
3.Ta xét từng trường hợp:
-TH1:\(\left\{{}\begin{matrix}x+1>0\\x-2< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x>-1\\x< 2\end{matrix}\right.\)\(\Rightarrow x\in\left\{0;1\right\}\)
-TH2:\(\left\{{}\begin{matrix}x+1< 0\\x-2>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x< -1\\x>2\end{matrix}\right.\)\(\Rightarrow x\in\varnothing\)
Vậy \(x\in\left\{0;1\right\}\)
4.\(B=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{9}{49}\right)^9=\left(\dfrac{3}{7}\right)^{21}:\left[\left(\dfrac{3}{7}\right)^2\right]^9=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{3}{7}\right)^{18}=\left(\dfrac{3}{7}\right)^3=\dfrac{27}{343}\)
\(\left(3-x\right)^3=-\dfrac{27}{64}\)
\(\left(3-x\right)^3=\left(\dfrac{-3}{4}\right)^3\)
\(=>3-x=\dfrac{-3}{4}\)
\(x=3-\dfrac{-3}{4}=\dfrac{12}{4}+\dfrac{3}{4}\)
\(x=\dfrac{15}{4}\)
________
\(\left(x-5\right)^3=\dfrac{1}{-27}\)
\(\left(x-5\right)^3=\left(\dfrac{-1}{3}\right)^3\)
\(=>x-5=\dfrac{-1}{3}\)
\(x=\dfrac{-1}{3}+5=\dfrac{-1}{3}+\dfrac{15}{3}\)
\(x=\dfrac{14}{3}\)
_____________
\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{27}{8}\)
\(\left(x-\dfrac{1}{2}\right)^3=\left(\dfrac{3}{2}\right)^3\)
\(=>x-\dfrac{1}{2}=\dfrac{3}{2}\)
\(x=\dfrac{3}{2}+\dfrac{1}{2}\)
\(x=2\)
________
\(\left(2x-1\right)^2=\dfrac{1}{4}\)
\(\left(2x-1\right)^2=\left(\dfrac{1}{2}\right)^2\) hoặc \(\left(2x-1\right)^2=\left(\dfrac{-1}{2}\right)^2\)
\(=>2x-1=\dfrac{1}{2}\) \(2x-1=\dfrac{-1}{2}\)
\(2x=\dfrac{1}{2}+1=\dfrac{1}{2}+\dfrac{2}{2}\) \(2x=\dfrac{-1}{2}+1=\dfrac{-1}{2}+\dfrac{2}{2}\)
\(2x=\dfrac{3}{2}\) \(2x=\dfrac{1}{2}\)
\(x=\dfrac{3}{2}:2=\dfrac{3}{2}.\dfrac{1}{2}\) \(x=\dfrac{1}{2}:2=\dfrac{1}{2}.\dfrac{1}{2}\)
\(x=\dfrac{3}{4}\) \(x=\dfrac{1}{4}\)
____________
\(\left(2-3x\right)^2=\dfrac{9}{4}\)
\(\left(2-3x\right)^2=\left(\dfrac{3}{2}\right)^2\) hoặc \(\left(2-3x\right)^2=\left(\dfrac{-3}{2}\right)^2\)
\(=>2-3x=\dfrac{3}{2}\) \(2-3x=\dfrac{-3}{2}\)
\(3x=2-\dfrac{3}{2}=\dfrac{4}{2}-\dfrac{3}{2}\) \(3x=2-\dfrac{-3}{2}=\dfrac{4}{2}+\dfrac{3}{2}\)
\(3x=\dfrac{1}{2}\) \(3x=\dfrac{7}{2}\)
\(x=\dfrac{1}{2}.\dfrac{1}{3}\) \(x=\dfrac{7}{2}.\dfrac{1}{3}\)
\(x=\dfrac{1}{6}\) \(x=\dfrac{7}{6}\)
______________
\(\left(1-\dfrac{2}{3}\right)^2=\dfrac{4}{9}\) -> Kiểm tra đề câu này
(3-x)3=(-\(\dfrac{3}{4}\))3
3-x=-\(\dfrac{3}{4}\)
x=3-(-\(\dfrac{3}{4}\))
x=\(\dfrac{15}{4}\)
\(2x-\dfrac{3}{4}=x+\dfrac{1}{3}\cdot\dfrac{-9}{2}\)
\(\Rightarrow2x-x=\dfrac{1}{3}\cdot\dfrac{-9}{2}+\dfrac{3}{4}\)
\(\Rightarrow x=-\dfrac{3}{2}+\dfrac{3}{4}\)
\(\Rightarrow x=-\dfrac{6}{4}+\dfrac{3}{4}=-\dfrac{3}{4}\)
#\(Toru\)
thanks