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a) \(\dfrac{1}{3}x-\dfrac{1}{2}=\dfrac{3}{4}x+\dfrac{1}{15}\)
\(\Rightarrow\dfrac{1}{3}x-\dfrac{3}{4}x=\dfrac{1}{2}+\dfrac{1}{15}\)
\(\Rightarrow\dfrac{4}{12}x-\dfrac{9}{12}x=\dfrac{15}{30}+\dfrac{2}{30}\)
\(\Rightarrow\dfrac{-5}{12}x=\dfrac{17}{30}\)
\(\Rightarrow x=\dfrac{-102}{75}\)
\(\left(x-\dfrac{2}{9}\right)^3=\left(\dfrac{2}{3}\right)^6\)
\(\Rightarrow\left(x-\dfrac{2}{9}\right)^3=\dfrac{64}{729}\)
\(\Rightarrow x-\dfrac{2}{9}=\dfrac{4}{9}\)
\(\Rightarrow x=\dfrac{2}{3}\)
Câu 3:
a: \(A=-\left|x-10\right|+2018< =2018\)
Dấu '=' xảy ra khi x=10
\(B=-\left(x+2\right)^2+1999< =1999\)
Dấu '=' xảy ra khi x=-2
b: \(A=\left(2x-8\right)^2+3>=3\)
Dấu '=' xảy ra khi x=4
\(B=\left|x^2-25\right|-2017>=-2017\)
Dấu '=' xảy ra khi x=5 hoặc x=-5
a)\(\dfrac{3}{10}\)-x=\(\dfrac{25}{30}\)-\(\dfrac{4}{30}\)
\(\dfrac{3}{10}-x=\dfrac{7}{10}\)
x = \(\dfrac{3}{10}-\dfrac{7}{10}\)
x=\(\dfrac{-4}{10}\)
b)\(\dfrac{-5}{8}+x=\dfrac{4}{9}-\dfrac{63}{9}\)
\(\dfrac{-5}{9}+x=\dfrac{-59}{9}\)
\(x=\dfrac{-59}{9}-\dfrac{-5}{9}\)
\(x=\dfrac{-64}{9}\)
c)=>2.18=(x-3).(x-3)
=>36=(x-3)\(^2\)
=>6\(^2\)=(x-3)\(^2\)
6= x-3
x=6+3=9
Cho \(A=\dfrac{\dfrac{-5}{8}.\dfrac{3}{7}+\dfrac{3}{7}.\dfrac{3}{4}+\dfrac{1}{7}.\dfrac{1}{2}+\dfrac{15}{8}}{a+\dfrac{5}{6}-\left(\dfrac{-1}{3}\right)}\)
a) Rút gọn A?
b) Tính A khi a=75%
c) Tìm a để A=50%
d) Tìm a thuộc Z để A là số nguyên.
e) Với a = bao nhiêu để A có giá trị bằng với giá trị của biểu thức:
\(B=\dfrac{\dfrac{2}{3}.\dfrac{15}{6}+\left(-0,5\right)^3}{\dfrac{1}{9}.6^2-5\dfrac{1}{3}}\)
Giải
a, Ta có:
\(A=\dfrac{\dfrac{-5}{8}.\dfrac{3}{7}+\dfrac{3}{7}.\dfrac{3}{4}+\dfrac{3}{7}.\dfrac{1}{6}+\dfrac{1}{8}.15}{a+\dfrac{5}{6}+\dfrac{1}{3}}\)
\(A=\dfrac{\dfrac{3}{7}.\left(\dfrac{-5}{8}+\dfrac{3}{4}+\dfrac{1}{6}\right)+\dfrac{1}{8}.15}{a+\dfrac{7}{6}}\)
\(A=\dfrac{\dfrac{3}{7}.\dfrac{7}{24}+\dfrac{1}{8}.15}{a+\dfrac{7}{6}}\)
\(A=\dfrac{\dfrac{1}{8}+\dfrac{1}{8}.15}{a+\dfrac{7}{6}}\)
\(A=\dfrac{\dfrac{1}{8}.\left(15+1\right)}{a+\dfrac{7}{6}}\)
\(A=\dfrac{2}{a+\dfrac{7}{6}}\)
b, Thay \(a=75\%\) vào \(A\), ta được:
\(A=\dfrac{2}{75\%+\dfrac{7}{6}}\)
\(A=\dfrac{2}{\dfrac{3}{4}+\dfrac{7}{6}}\)
\(\Rightarrow A=\dfrac{23}{12}\)
c, Ta có: \(\dfrac{2}{a+\dfrac{7}{6}}=50\%\)
\(\dfrac{2}{a+\dfrac{7}{6}}=\dfrac{1}{2}\)
\(\dfrac{2}{a+\dfrac{7}{6}}=\dfrac{2}{4}\)
\(\Rightarrow a+\dfrac{7}{6}=4\)
\(\Rightarrow a=\dfrac{17}{6}\)
d, Để \(A\in Z\Rightarrow2⋮a+\dfrac{7}{6}\)
\(\Rightarrow a+\dfrac{7}{6}\in\left\{\pm1;\pm2\right\}\)
\(\circledast,a+\dfrac{7}{6}=1\Rightarrow a=\dfrac{-1}{6}\)
\(\circledast,a+\dfrac{7}{6}=-1\Rightarrow a=\dfrac{-13}{6}\)
\(\circledast,a+\dfrac{7}{6}=2+\Rightarrow a=\dfrac{5}{6}\)
\(\circledast,a+\dfrac{7}{6}=-2\Rightarrow a=\dfrac{-19}{6}\)
\(a\in\varnothing\) khi \(A\in Z\)
e, Ta có:
\(B=\dfrac{5}{3}+\dfrac{-1}{8}\Rightarrow B=\dfrac{37}{24}\)
\(\Rightarrow\dfrac{2}{a+\dfrac{7}{6}}=\dfrac{37}{24}\)
\(a+\dfrac{7}{6}=\dfrac{37}{24}.2\)
\(a+\dfrac{7}{6}=\dfrac{37}{12}\)
\(\Rightarrow a=\dfrac{23}{12}\)
Chúc bạn học thiệt giỏi nha!!! 
)
Áp dụng tính chất phân phối, rồi tính giá trị biểu thức.
Chẳng hạn,
Với , thì
ĐS. ; C = 0.
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a) Để \(A=\dfrac{5}{\left(x-3\right)^2+1}\) đạt giá trị lớn nhất
\(\Leftrightarrow\left(x-3\right)^2+1\) phải nhỏ nhất
Mà \(\left(x-3\right)^2\ge0\Leftrightarrow\left(x-3\right)^2+1\ge1\)
\(\Rightarrow A_{max}=\dfrac{5}{\left(x-3\right)^2+1}=\dfrac{5}{1}=5\)
Dấu "=" xảy ra \(\Leftrightarrow\left(x-3\right)^2=0\Leftrightarrow x-3=0\Rightarrow x=3\)
Vậy \(A_{max}=5\) tại \(x=3\)
b) Để \(B=\dfrac{4}{\left|x-2\right|+2}\) đạt giá trị lớn nhất
\(\Leftrightarrow\left|x-2\right|+2\) phải nhỏ nhất
Mà \(\left|x-2\right|\ge0\Leftrightarrow\left|x-2\right|+2\ge2\)
\(\Rightarrow B_{max}=\dfrac{4}{\left|x-2\right|+2}=\dfrac{4}{2}=2\)
Dấu "=" xảy ra \(\Leftrightarrow\left|x-2\right|=0\Leftrightarrow x-2=0\Rightarrow x=2\)
Vậy \(B_{max}=2\) tại \(x=2\)
\(A=-1,6:\left(1+\dfrac{2}{3}\right)\)
\(A=\dfrac{-16}{10}:\dfrac{5}{3}\)
\(A=\dfrac{-8}{5}.\dfrac{3}{5}\)
\(A=\dfrac{-24}{25}\)
\(B=1,4.\dfrac{15}{49}-\left(\dfrac{4}{5}+\dfrac{2}{3}\right):2\dfrac{1}{5}\)
\(B=\dfrac{14}{10}.\dfrac{15}{49}-\left(\dfrac{4}{5}+\dfrac{2}{3}\right):\dfrac{11}{5}\)
\(B=\dfrac{14}{10}.\dfrac{15}{49}-\dfrac{22}{15}:\dfrac{11}{5}\)
\(B=\dfrac{3}{7}-\dfrac{22}{15}:\dfrac{11}{5}\)
\(B=\dfrac{3}{7}-\dfrac{2}{3}\)
\(B=\dfrac{-5}{21}\)
\(A=-1,6:\left(1+\dfrac{2}{3}\right)\)
\(A=\dfrac{-8}{5}:\left(1+\dfrac{2}{3}\right)\)
\(A=\dfrac{-8}{5}:\dfrac{5}{3}\)
\(A=\dfrac{-24}{25}\)
\(B=1,4.\dfrac{15}{49}-\left(\dfrac{4}{5}+\dfrac{2}{3}\right):2\dfrac{1}{5}\)
\(B=\dfrac{7}{5}.\dfrac{15}{49}-\left(\dfrac{4}{5}+\dfrac{2}{3}\right):\dfrac{11}{5}\)
\(B=\dfrac{7}{5}.\dfrac{15}{49}-\dfrac{22}{15}:\dfrac{11}{5}\)
\(B=\dfrac{3}{7}-\dfrac{2}{3}\)
\(B=\dfrac{-5}{21}\)
a) Ta có: \(\left|2x-\dfrac{1}{3}\right|\ge0\forall x\)
\(\Leftrightarrow\left|2x-\dfrac{1}{3}\right|-\dfrac{7}{4}\ge-\dfrac{7}{4}\forall x\)
Dấu '=' xảy ra khi \(2x=\dfrac{1}{3}\)
hay \(x=\dfrac{1}{6}\)
Vậy: \(A_{min}=-\dfrac{7}{4}\) khi \(x=\dfrac{1}{6}\)
b) Ta có: \(\dfrac{1}{3}\left|x-2\right|\ge0\forall x\)
\(\left|3-\dfrac{1}{2}y\right|\ge0\forall y\)
Do đó: \(\dfrac{1}{3}\left|x-2\right|+\left|3-\dfrac{1}{2}y\right|\ge0\forall x,y\)
\(\Leftrightarrow\dfrac{1}{3}\left|x-2\right|+\left|3-\dfrac{1}{2}y\right|+4\ge4\forall x,y\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-2=0\\3-\dfrac{1}{2}y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=6\end{matrix}\right.\)
Vậy: \(B_{min}=4\) khi x=2 và y=6
Cảm ơn nhiều nha !