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\(\dfrac{5}{6}x-\dfrac{3}{4}=\dfrac{-1}{4}+\dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{5}{6}x=\dfrac{7}{6}\)
\(\Rightarrow x=\dfrac{7}{5}\)
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: \(\dfrac{2.75}{x}=\dfrac{0.4}{1.5}=\dfrac{4}{15}\)
\(\Leftrightarrow x=\dfrac{11}{4}\cdot\dfrac{15}{4}=\dfrac{165}{16}\)
b: \(3\dfrac{1}{2}:\left(2x-3\right)=\dfrac{-3}{4}:0.2\)
\(\Leftrightarrow\dfrac{7}{2}:\left(2x-3\right)=\dfrac{-3}{4}:\dfrac{1}{5}=\dfrac{-15}{4}\)
\(\Leftrightarrow2x-3=\dfrac{7}{2}:\dfrac{-15}{4}=\dfrac{-7}{2}\cdot\dfrac{4}{15}=\dfrac{-28}{30}=\dfrac{-14}{15}\)
=>2x=-14/15+3=45/45-14/15=31/45
=>x=31/90
c: \(\dfrac{3x+2}{27}=\dfrac{3}{3x+2}\)
\(\Leftrightarrow\left(3x+2\right)^2=81\)
=>3x+2=9 hoặc 3x+2=-9
=>3x=7 hoặc 3x=-11
=>x=7/3 hoặc x=-11/3
d: \(\dfrac{5-x}{4}=\dfrac{2x+3}{2}\)
=>10-2x=8x+12
=>-10x=2
hay x=-1/5
a)
\(\left\{{}\begin{matrix}\left(4x-1\right)^4\ge0\\\left|2x-3y\right|\ge0\end{matrix}\right.\) \(\Rightarrow A\ge25,6\) tự tìm cận
không có Max
b) giống vậy
c) \(\left\{{}\begin{matrix}\left(x-3\right)^2\ge0\Rightarrow-\left(x-3\right)^2\le0\\\left|4x-3y\right|\ge0\Rightarrow-\left|4x-3y\right|\le0\end{matrix}\right.\)
\(C\le40,5\) tự tìm cận
không có GTNN
a/ \(\dfrac{1}{3}-\dfrac{2}{5}+3x=\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{-1}{15}+3x=\dfrac{3}{4}\)
\(\Leftrightarrow3x=\dfrac{49}{60}\)
\(\Leftrightarrow x=\dfrac{49}{180}\)
Vậy....
b/ \(\dfrac{3}{2}-1+4x=\dfrac{2}{3}-7x\)
\(\Leftrightarrow\dfrac{1}{2}+4x=\dfrac{2}{3}-7x\)
\(\Leftrightarrow4x+7x=\dfrac{2}{3}-\dfrac{1}{2}\)
\(\Leftrightarrow11x=\dfrac{1}{6}\)
\(\Leftrightarrow x=\dfrac{1}{66}\)
Vậy....
c/ \(2\left(\dfrac{3}{4}-5x\right)=\dfrac{4}{5}-3x\)
\(\Leftrightarrow\dfrac{3}{2}-10x=\dfrac{4}{5}-3x\)
\(\Leftrightarrow-10x+3x=\dfrac{4}{5}-\dfrac{3}{2}\)
\(\Leftrightarrow-7x=-\dfrac{7}{10}\)
\(\Leftrightarrow x=-\dfrac{1}{10}\)
Vậy .....
d/ \(4\left(\dfrac{1}{2}-x\right)-5\left(x-\dfrac{3}{10}\right)=\dfrac{7}{4}\)
\(\Leftrightarrow2-4x-5x-\dfrac{3}{2}=\dfrac{7}{4}\)
\(\Leftrightarrow2+\left(-4x\right)+\left(-5x\right)+\left(\dfrac{-3}{2}\right)=\dfrac{7}{4}\)
\(\Leftrightarrow-9x+\dfrac{1}{2}=\dfrac{7}{4}\)
\(\Leftrightarrow-9x=\dfrac{5}{4}\)
\(\Leftrightarrow x=-\dfrac{5}{36}\)
a)
\(\left(3x+\dfrac{1}{3}\right)\left(x-\dfrac{1}{2}\right)=0\\ \Rightarrow\left[{}\begin{matrix}3x+\dfrac{1}{3}=0\\x-\dfrac{1}{2}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{9}\\x=\dfrac{1}{2}\end{matrix}\right.\)
b)
\(\left(x-\dfrac{3}{2}\right)\left(2x+1\right)>0\\ \Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-\dfrac{3}{2}>0\\2x+1>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-\dfrac{3}{2}< 0\\2x+1< 0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>\dfrac{3}{2}\\x>-\dfrac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x< \dfrac{3}{2}\\x< -\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x>\dfrac{3}{2}\\x< -\dfrac{1}{2}\end{matrix}\right.\)
1. Tìm x:
a) \(\left(x+36\right)^2=1936\Leftrightarrow x+36=\pm44.\) Vậy x = 8 hoặc x = -80
b) \(\left(\dfrac{3}{5}\right)^{x+2}=\dfrac{81}{625}\Leftrightarrow\left(\dfrac{3}{5}\right)^{x+2}=\left(\dfrac{3}{5}\right)^4\Leftrightarrow x+2=4\Leftrightarrow x=2\)
c) Xem lại đề
d) \(\left(\dfrac{9}{16}\right)^{x-5}=\left(\dfrac{4}{3}\right)^4\Leftrightarrow\left(\dfrac{3}{4}\right)^{2\left(x-5\right)}=\left(\dfrac{3}{4}\right)^{-4}\Leftrightarrow2\left(x-5\right)=-4\Leftrightarrow x=3\)
e) \(\left(\dfrac{3}{5}\right)^x.\left(\dfrac{125}{27}\right)^x=\dfrac{81}{625}\Leftrightarrow\left(\dfrac{3}{5}.\dfrac{125}{27}\right)^x=\left(\dfrac{3}{5}\right)^4\Leftrightarrow\left(\dfrac{5}{3}\right)^{2x}=\left(\dfrac{5}{3}\right)^{-4}\Leftrightarrow2x=-4\) Vậy x = -2
3. Tính giá trị của biểu thức:
\(A=\left\{-\left[\left(\dfrac{1}{x}\right)^2\right]^3\right\}^5.\left\{-\left[\left(-x\right)^5\right]^2\right\}^3\) \(\left(x\notin0\right)\)
\(=\left\{-\left[-\dfrac{1}{x^2}\right]^3\right\}^5.\left\{-\left[-\left(-x\right)^5\right]^2\right\}^3=\left\{-\left[-\dfrac{1}{x^6}\right]\right\}^5.\left\{-\left[x^5\right]^2\right\}^3\)
\(=\left\{\dfrac{1}{x^6}\right\}^5.\left\{-x^{10}\right\}^3=\dfrac{1}{x^{30}}.\left(-x^{30}\right)=-1\)
c) +)Điểm A ( 1;9) => x = 1 ; y = 9
Thay x = 1 vào y = 4x+5 , ta có:
y = 4.1+5
y = 4+5
y = 9
Vậy điểm A ( 1;9 ) thuộc đồ thị hàm số y = 4x +5
+) Điểm B ( -2;3 ) => x = -2 ; y = 3
Thay x = -2 vào y = 4x +5 , ta có:
y = 4.(-2) + 5
y = (-8) + 5
y = (-3)
Vậy điểm B ( -2;3) không thuộc đồ thị hàm số y = 4x+5
....Các câu khác tương tự....> . <...
a: \(=-\dfrac{1}{15}x^6y\)
b: \(=\dfrac{4}{5}ab^5\cdot2x^3y\cdot\left(-y\right)=-\dfrac{8}{5}ab^5\cdot x^3y^2\)
c: \(=-16\cdot\dfrac{3}{4}v^3\cdot\dfrac{-2}{5}uv=\dfrac{24}{5}v^4u\)
d: \(=8\cdot\left(-64\right)\cdot5\cdot u^2v^2\cdot\left(-27\right)v^3=69120u^2v^5\)
e: \(=-10y\cdot8y^3z^3\cdot25z^2=-2000y^4z^5\)
1.Tính
a.\(\dfrac{7}{23}\left[(-\dfrac{8}{6})-\dfrac{45}{18}\right]=\dfrac{7}{23}.-\dfrac{12}{6}=-\dfrac{7}{6}\)
b.\(\dfrac{1}{5}\div\dfrac{1}{10}-\dfrac{1}{3}(\dfrac{6}{5}-\dfrac{9}{4})=2-(-\dfrac{7}{20})=\dfrac{47}{20}\)
c.\(\dfrac{3}{5}.(-\dfrac{8}{3})-\dfrac{3}{5}\div(-6)=-\dfrac{3}{2}\)
d.\(\dfrac{1}{2}.(\dfrac{4}{3}+\dfrac{2}{5})-\dfrac{3}{4}.(\dfrac{8}{9}+\dfrac{16}{3})=-\dfrac{19}{5}\)
e.\(\dfrac{6}{7}\div(\dfrac{3}{26}-\dfrac{3}{13})+\dfrac{6}{7}.(\dfrac{1}{10}-\dfrac{8}{5})=-\dfrac{61}{7}\)
Bài 2
a.\(1^2_5x+\dfrac{3}{7}=\dfrac{4}{5}\)
\(x=\dfrac{13}{49}\)
b.\(\left|x-1,5\right|=2\)
Xảy ra 2 trường hợp
TH1
\(x-1,5=2\)
\(x=3,5\)
TH2
\(x-1,5=-2\)
\(x=-0,5\)
Vậy \(x=3,5\) hoặc \(x=-0,5\) .
Ngại làm quá trời ơi,lần sau bn tách ra nhá làm vậy mỏi tay quá.
a.\(\left(3x-2\right)^2=16\)
Ta có: \(\left(3x-2\right)^2=16\)
\(\Rightarrow\left(3x-2\right)^2=\left(4\right)^2\)
\(\Rightarrow3x-2=4\)
\(\Rightarrow3x=6\)
\(\Rightarrow x=2\)
b. \(\left(\dfrac{4}{5}x-\dfrac{3}{4}\right)^3=\dfrac{-8}{125}\)
\(\Rightarrow\left(\dfrac{4}{5}x-\dfrac{3}{4}\right)^3=\left(\dfrac{-2}{5}\right)^3\)
\(\Rightarrow\dfrac{4}{5}x-\dfrac{3}{4}=\dfrac{-2}{5}^{ }\)
\(\Rightarrow\dfrac{4}{5}x-=\dfrac{7}{20}\)
\(\Rightarrow x=\dfrac{7}{16}\)