toi ko co the bt day nh vau ko dau

\(\left(x-\frac{1}{2}\right)^2=0\)

\(\Leftrightarrow\left(x-\frac{1}{2}\right)^2=0^2\)

\(\Leftrightarrow x-\frac{1}{2}=0\)

\(\Leftrightarrow x=\frac{1}{2}\)

Vậy x = 1/2

\(\left(x-2\right)^2=1\)

\(\Leftrightarrow\left(x-2\right)^2=1^2\)

\(\Leftrightarrow x-2=1\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=1\\x-2=-1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}}\)

Vậy x = 3 hoặc x = 1

\(\left(2x-1\right)^3=-8\)

\(\Leftrightarrow\left(2x-1\right)^3=\left(-2\right)^3\)

\(\Leftrightarrow2x-1=-2\)

<=> 2x = -1

<=> x = -0,5

Vậy x = -0,5

\(\left(x-\frac{1}{2}\right)^2=0\)

\(x-\frac{1}{2}=0\)

\(x=\frac{1}{2}\)

\(\left(x-2\right)^2=1\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=1\\x-2=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1+2\\x=-1+2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}\)

Vậy\(x\in\left\{3;1\right\}\)
\(\left(2x-1\right)^3=-8\)

\(\left(2x-1\right)^3=\left(-2\right)^3\)

\(2x-1=-2\)

\(2x=\left(-2\right)+1\)

\(2x=-1\)

\(x=-1\times2\)

\(x=-2\)

\(x\left(\frac{1}{2}\right)^2=\frac{1}{16}\)

\(x\left(\frac{1}{2}\right)^2=\left(\frac{1}{4}\right)^2\)

\(\Leftrightarrow\orbr{\begin{cases}x\frac{1}{2}=\frac{1}{4}\\x\frac{1}{2}=-\frac{1}{4}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{4}:\frac{1}{2}\\x=-\frac{1}{4}:\frac{1}{2}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{1}{2}\end{cases}}}\)

\(49\times7^x=2401\)

   \(7^x=2401:49\)

\(7^x=49\)

\(7^x=7^2\)

\(\Rightarrow x=2\)

\(2x^x-15=17\)

\(2x^x=17+15\)

\(2x^x=32\)

\(2x^x=2^5\)

\(\Rightarrow x\times x=5\)

\(\Rightarrow x^2=5\)

\(2^x+3\times2^x=32\)

\(2^x\times\left(3+1\right)=32\)

\(2^x\times4=32\)

\(2^x=32:4\)

\(2^x=8\)

\(2^x=2^3\)

\(\Rightarrow x=3\)

\(2x\left(x-\frac{1}{7}\right)=0\)

\(x\left(x-\frac{1}{7}\right)=0:2\)

\(x\left(x-\frac{1}{7}\right)=0\)

\(x\times x-x\times\frac{1}{7}=0\)

\(x^2-x\frac{1}{7}=0\)

\(x-x=0:\frac{1}{7}\)

\(x-x=0\)

\(\Rightarrow x=0\)

Đặt A=\(\frac{1}{3}.5+\frac{1}{5}.7+...+\frac{1}{97}.99\)

=>A=\(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)

=>2A=\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\)

=>2A=\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)

=>2A=\(\frac{1}{3}-\frac{1}{99}=\frac{33}{99}-\frac{1}{99}=\frac{32}{99}\)

=>A=\(\frac{32}{99}:2=\frac{32}{99}.\frac{1}{2}=\frac{32}{198}=\frac{16}{99}\)

BẠN GHI ĐỀ LẠI CÓ DẤU PHẨY ĐI Ạ

5 . y . \(\frac{1}{2}\). x³y(\(\frac{-1}{3}\).x².y)³= \(\frac{5}{2}\)x³y² \(\frac{-1}{27}\) x⁶y³= \(\frac{-5}{54}\)x⁹y⁵

Hệ số \(\frac{-5}{54}\) 

Phần biến : x⁹y⁵

Bậc : 14

Chúc bạn học tốt !!!

Đề bài yêu cầu j vậy?

mũ 2 tui ko bt

chịu bạn ơi

Ta có : \(\frac{1}{2^2}<\frac{1}{1.2}\)

            \(\frac{1}{2^3}<\frac{1}{2.3}\)

            \(\frac{1}{2^4}<\frac{1}{3.4}\)

             ..........

             \(\frac{1}{2^n}<\frac{1}{\left(n-1\right).n}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+....+\frac{1}{2^n}<\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}=1-\frac{1}{n}\)

Mà \(1-\frac{1}{n}<1\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+.....+\frac{1}{2^n}<1\left(đpcm\right)\)

cảm ơn bạn nha mình tích cho bạn rùi đấy

a/ \(\left(-2x^2y\right)5xy^4\)

\(=-10x^3y^5\)

tự lm tiếp, rất dễ

a) Ta có: \(\left(-2x^2y\right)\cdot\left(5xy^4\right)\)

\(=\left(-2\cdot5\right)\cdot\left(x^2\cdot x\right)\cdot\left(y\cdot y^4\right)\)

\(=-10x^3y^5\)

b) Ta có: \(\left(\dfrac{27}{10}x^4y^2\right)\cdot\left(\dfrac{5}{9}xy\right)\)

\(=\left(\dfrac{27}{10}\cdot\dfrac{5}{9}\right)\cdot\left(x^4\cdot x\right)\cdot\left(y^2\cdot y\right)\)

\(=\dfrac{3}{2}x^5y^3\)

c) Ta có: \(\left(\dfrac{1}{3}x^3y\right)\cdot\left(-xy\right)^2\)

\(=\dfrac{1}{3}x^3y\cdot x^2y^2\)

\(=\dfrac{1}{3}x^5y^3\)

Bài 6 :

a) \(\dfrac{625}{5^n}=5\Rightarrow\dfrac{5^4}{5^n}=5\Rightarrow5^{4-n}=5^1\Rightarrow4-n=1\Rightarrow n=3\)

b) \(\dfrac{\left(-3\right)^n}{27}=-9\Rightarrow\dfrac{\left(-3\right)^n}{\left(-3\right)^3}=\left(-3\right)^2\Rightarrow\left(-3\right)^{n-3}=\left(-3\right)^2\Rightarrow n-3=2\Rightarrow n=5\)

c) \(3^n.2^n=36\Rightarrow\left(2.3\right)^n=6^2\Rightarrow\left(6\right)^n=6^2\Rightarrow n=6\)

d) \(25^{2n}:5^n=125^2\Rightarrow\left(5^2\right)^{2n}:5^n=\left(5^3\right)^2\Rightarrow5^{4n}:5^n=5^6\Rightarrow\Rightarrow5^{3n}=5^6\Rightarrow3n=6\Rightarrow n=3\)

Bài 7 :

a) \(3^x+3^{x+2}=9^{17}+27^{12}\)

\(\Rightarrow3^x\left(1+3^2\right)=\left(3^2\right)^{17}+\left(3^3\right)^{12}\)

\(\Rightarrow10.3^x=3^{34}+3^{36}\)

\(\Rightarrow10.3^x=3^{34}\left(1+3^2\right)=10.3^{34}\)

\(\Rightarrow3^x=3^{34}\Rightarrow x=34\)

b) \(5^{x+1}-5^x=100.25^{29}\Rightarrow5^x\left(5-1\right)=4.5^2.\left(5^2\right)^{29}\)

\(\Rightarrow4.5^x=4.25^{2.29+2}=4.5^{60}\)

\(\Rightarrow5^x=5^{60}\Rightarrow x=60\)

c) Bài C bạn xem lại đề

d) \(\dfrac{3}{2.4^x}+\dfrac{5}{3.4^{x+2}}=\dfrac{3}{2.4^8}+\dfrac{5}{3.4^{10}}\)

\(\Rightarrow\dfrac{3}{2.4^x}-\dfrac{3}{2.4^8}+\dfrac{5}{3.4^{x+2}}-\dfrac{5}{3.4^{10}}=0\)

\(\Rightarrow\dfrac{3}{2}\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)+\dfrac{5}{3.4^2}\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)=0\)

\(\Rightarrow\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)\left(\dfrac{3}{2}+\dfrac{5}{3.4^2}\right)=0\)

\(\Rightarrow\dfrac{1}{4^x}-\dfrac{1}{4^8}=0\)

\(\Rightarrow\dfrac{4^8-4^x}{4^{x+8}}=0\Rightarrow4^8-4^x=0\left(4^{x+8}>0\right)\Rightarrow4^x=4^8\Rightarrow x=8\)

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