Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a,\Rightarrow\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}=\left(\dfrac{1}{3}\right)^3\\ \Rightarrow x-\dfrac{1}{2}=\dfrac{1}{3}\Rightarrow x=\dfrac{5}{6}\\ b,\Rightarrow\left(\dfrac{3}{2}\right)^{2x-1}:\left(\dfrac{3}{2}\right)^9=\left(\dfrac{3}{2}\right)^4\\ \Rightarrow2x-1-9=4\\ \Rightarrow2x=14\Rightarrow x=7\\ c,\Rightarrow2^{x-1}+2^{x+2}=9\cdot2^5\\ \Rightarrow2^{x-1}\left(1+2^3\right)=9\cdot2^5\\ \Rightarrow2^{x-1}\cdot9=9\cdot2^5\\ \Rightarrow2^{x-1}=2^5\Rightarrow x-1=5\Rightarrow x=6\\ d,\Rightarrow\left(2x+1\right)^2=12+69=81\\ \Rightarrow\left[{}\begin{matrix}2x+1=9\\2x+1=-9\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-5\end{matrix}\right.\)
a: \(\left(\dfrac{5}{9}-\dfrac{\sqrt{9}}{12}\right):\dfrac{3}{4}+\dfrac{11}{3}:\dfrac{3}{4}\)
\(=\left(\dfrac{5}{9}-\dfrac{3}{12}\right)\cdot\dfrac{4}{3}+\dfrac{11}{3}\cdot\dfrac{4}{3}\)
\(=\left(\dfrac{5}{9}-\dfrac{1}{4}+\dfrac{11}{3}\right)\cdot\dfrac{4}{3}\)
\(=\dfrac{20-9+132}{36}\cdot\dfrac{4}{3}\)
\(=\dfrac{143}{3}\cdot\dfrac{1}{9}=\dfrac{143}{27}\)
b: \(\left(0.\left(3\right)+\dfrac{\left|-2\right|}{3}\right):\dfrac{\sqrt{25}}{4}-\left(2^3+3^2\right)^0\)
\(=\left(\dfrac{1}{3}+\dfrac{2}{3}\right)\cdot\dfrac{4}{5}-1\)
\(=\dfrac{4}{5}-1=-\dfrac{1}{5}\)
a: \(2\left|3-2x\right|+\dfrac{1}{2}=\dfrac{5}{2}\)
=>\(2\left|2x-3\right|=2\)
=>|2x-3|=1
=>\(\left[{}\begin{matrix}2x-3=1\\2x-3=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=4\\2x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\end{matrix}\right.\)
b: \(x^2\left(2^x-6\right)-2x^3=0\)
=>\(x^2\left(2^x-6-2x\right)=0\)
=>\(\left[{}\begin{matrix}x^2=0\\2^x-6-2x=0\end{matrix}\right.\Leftrightarrow x=0\)
a) \(A=2x^2-\dfrac{1}{3}y\)
A= \(\left(2-\dfrac{1}{3}\right)\)\(x^2y\)
A=\(\dfrac{5}{3}\)\(x^2y\)
Tại \(x=2;y=9\) ta có
A=\(\dfrac{5}{3}\).(2)\(^2\).9 = \(\dfrac{5}{3}\).4 .9 = 60
Vậy tại \(x=2;y=9\) biểu thức A= 60
b) P=\(2x^2+3xy+y^2\) (\(y^2\) là 1\(y^2\) nha bạn)
P=\(\left(2+3+1\right)\left(x^2.x\right)\left(y.y^2\right)\)
P= 6\(x^3y^3\)
Tại \(x=-\dfrac{1}{2};y=\dfrac{2}{3}\) ta có
P= 6.\(\left(-\dfrac{1}{2}\right)^3.\left(\dfrac{2}{3}\right)^3\) = 6.\(\left(-\dfrac{1}{8}\right).\dfrac{8}{27}\) = \(-\dfrac{2}{9}\)
Vậy tại \(x=-\dfrac{1}{2};y=\dfrac{2}{3}\) biểu thức P= \(-\dfrac{2}{9}\)
c)\(\left(-\dfrac{1}{2}xy^2\right).\left(\dfrac{2}{3}x^3\right)\)
=\(\left((-\dfrac{1}{2}).\dfrac{2}{3}\right)\left(x.x^3\right).y^2\)
=\(-\dfrac{1}{3}\)\(x^4y^2\)
Tại \(x=2;y=\dfrac{1}{4}\)ta có
\(-\dfrac{1}{3}\).\(\left(2\right)^4.\left(\dfrac{1}{4}\right)^2=-\dfrac{1}{3}.16.\dfrac{1}{16}=-\dfrac{1}{3}\)
\(\)Vậy \(x=2;y=\dfrac{1}{4}\) biểu thức \(\left(-\dfrac{1}{2}xy^2\right).\left(\dfrac{2}{3}x^3\right)\)= \(-\dfrac{1}{3}\)
CHÚC BẠN HỌC TỐT NHA
a: \(\left(0.5\right)^3\cdot2^3=1\)
b: \(\left(0.25\right)^2\cdot16=1\)
c: \(\left(\dfrac{3}{5}\right)^3:\left(-\dfrac{27}{1000}\right)=\dfrac{3^3}{5^3}\cdot\dfrac{-1000}{27}=\dfrac{-1000}{125}=-8\)
Bài 1:
a: \(A=\left(-\dfrac{1}{5}\right)^{33}:\left(-\dfrac{1}{5}\right)^{32}=\dfrac{-1}{5}\)
c: \(C=\dfrac{2^{12}\cdot3^{10}+3^9\cdot2^9\cdot2^3\cdot3\cdot5}{2^{12}\cdot3^{12}+2^{11}\cdot3^{11}}\)
\(=\dfrac{2^{12}\cdot3^{10}\left(1+5\right)}{2^{11}\cdot3^{11}\cdot7}=\dfrac{2}{3}\cdot\dfrac{6}{7}=\dfrac{12}{21}=\dfrac{4}{7}\)
1) Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Ta có \(\dfrac{3a+5b}{3a-5b}=\dfrac{3bk+5b}{3bk-5b}=\dfrac{b\left(3k+5\right)}{b\left(3k-5\right)}=\dfrac{3k+5}{3k-5}\) (1)
\(\dfrac{3c+5d}{3c-5d}=\dfrac{3dk+5d}{3dk-5d}=\dfrac{d\left(3k+5\right)}{d\left(3k-5\right)}=\dfrac{3k+5}{3k-5}\) (2)
Từ (1) và (2) \(\Rightarrow\dfrac{3a+5b}{3a-5b}=\dfrac{3c+5d}{3c-5d}\)
2) Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=q\Rightarrow\left\{{}\begin{matrix}a=bq\\c=dq\end{matrix}\right.\)
Ta có: \(\left(\dfrac{a+b}{c+d}\right)^2=\left(\dfrac{bq+b}{dq+d}\right)^2=\left[\dfrac{b\left(q+1\right)}{d\left(q+1\right)}\right]^2=\dfrac{b}{d}\) (1)
\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{\left(bq\right)^2+b^2}{\left(dq\right)^2+d^2}=\dfrac{b^2.q^2+b^2}{d^2.q^2+d^2}=\dfrac{b^2\left(q^2+1\right)}{d^2\left(q^2+1\right)}=\dfrac{b^2}{d^2}=\dfrac{b}{d}\) (2)
Từ (1) và (2) \(\Rightarrow\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)
\(A=\dfrac{\left(-2\right)^0+1^{2017}+\left(\dfrac{-1}{3}\right)^8.3^8}{2^{15}}=\dfrac{3}{2^{15}}\left(1\right)\)
\(B=\dfrac{6^2}{2^{16}}\left(2\right)\)
\(\left(1\right);\left(2\right)\Rightarrow\dfrac{A}{B}=\dfrac{\dfrac{3}{2^{15}}}{\dfrac{6^2}{2^{16}}}=\dfrac{1}{6}\)