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\(18^{20}.45^5.5^{25}.8^{10}\)
\(=3^{40}.2^{20}.5^5.3^{10}.5^{25}.2^{30}\)
\(=3^{50}.2^{50}.5^{30}\)
\(=6^{50}.5^{30}\)
\(=\left(6^5\right)^{10}.\left(5^3\right)^{10}\)
\(=\left(6^5.5^3\right)^{10}\)
\(\left(x^2y\right)^5.\left(x^2.y^2\right)^7.\left(x.y\right)^6.x^3\)
\(=x^{10}.y^5.x^{14}.y^{14}.x^6.y^3.x^3\)
\(=x^{33}.y^{22}\)
\(=\left(x^3\right)^{11}.\left(y^2\right)^{11}\)
\(=\left(x^3.y^2\right)^{11}\)
\(2^7.3^8.4^9.9^8\)
\(=2^7.3^8.2^{18}.3^{16}\)
\(=2^{25}.3^{24}\)( mk chỉ làm được đến thế thôi )
Tham khảo nhé~
a) \(18^{20}.45^5.5^{25}.8^{10}\)
\(=\left(2.3^2\right)^{20}.\left(3^2.5\right)^5.5^{25}.\left(2^3\right)^{10}\)
\(=2^{20}.3^{40}.3^{10}.5^5.5^{25}.2^{30}\)
\(=2^{50}.3^{50}.5^{30}\)
\(=6^{50}.5^{30}\)
\(=\left(6^5\right)^{10}.\left(5^3\right)^{10}\)
\(=7776^{10}.125^{10}\)
\(=972000^{10}\)
b ) \(\left(x^2y\right)^5.\left(x^2.y^2\right)^7.\left(xy\right)^6.x^3\)
\(=x^{10}.y^5.x^{14}.y^{14}.x^6.y^6.x^3\)
\(=x^{33}.y^{25}\)
\(=x^{25}.y^{25}.x^8\)
\(=...\)
c) \(2^7.3^8.4^9.9^8\)
\(=2^7.3^8.\left(2^2\right)^9.\left(3^2\right)^8\)
\(=2^7.3^8.2^{18}.3^{16}\)
\(=2^{25}.3^{24}\)
\(=...\)( Câu c này hình như đề bài sai sót . Không chuyển thành lũy thừa được )
Bài 1:
a) b) c) sẽ có bạn giải cho em thôi vì nó dễ tính tay cũng đc
d) \(\frac{4}{2.5}+\frac{4}{5.8}+...+\frac{4}{23.26}\)
\(=\frac{4}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{23.26}\right)\)
\(=\frac{4}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{23}-\frac{1}{26}\right)\)
\(=\frac{4}{3}.\left(\frac{1}{2}-\frac{1}{26}\right)\)
\(=\frac{4}{3}.\frac{6}{13}\)
\(=\frac{8}{13}\)
Bài 2:
a) b) c)
d)\(|\frac{5}{8}x+\frac{6}{7}|-\frac{4}{7}=\frac{10}{7}\)
\(\Leftrightarrow|\frac{5}{8}x+\frac{6}{7}|=2\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{5}{8}x+\frac{6}{7}=2\\\frac{5}{8}x+\frac{6}{7}=-2\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}\frac{5}{8}x=\frac{8}{7}\\\frac{5}{8}x=\frac{-20}{7}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{64}{35}\\x=\frac{-32}{7}\end{cases}}}\)
Vậy \(x\in\left\{\frac{64}{35};\frac{-32}{7}\right\}\)
Bài 1 :
a) \(\left(\frac{2}{5}-\frac{5}{8}\right):\frac{11}{30}+\frac{1}{8}\)
\(=\frac{-9}{40}:\frac{11}{30}+\frac{1}{8}\)
\(=\frac{-27}{44}+\frac{1}{8}\)
\(=\frac{-43}{88}\)
\(1\frac{13}{15}.0,75-\left(\frac{8}{15}+25\%\right).\frac{24}{47}-3\frac{12}{13}:3\)
\(=\frac{28}{15}.\frac{3}{4}-\left(\frac{8}{15}+\frac{1}{4}\right).\frac{24}{47}-\frac{51}{13}:3\)
\(=\frac{7}{5}-\frac{47}{60}.\frac{24}{47}-\frac{17}{13}\)
\(=\frac{7}{5}-\frac{2}{5}-\frac{17}{13}\)
\(=\frac{-4}{13}\)
\(4\frac{1}{3}.\left(\frac{1}{6}-\frac{1}{2}\right)\le x\le\frac{2}{3}.\left(\frac{1}{3}-\frac{1}{2}-\frac{3}{4}\right)\)
\(\Leftrightarrow\frac{13}{3}.\frac{-1}{3}\le x\le\frac{2}{3}.\frac{-11}{12}\)
\(\Leftrightarrow\frac{-13}{9}\le x\le\frac{-11}{18}\)
\(\Leftrightarrow x=-1\)
a) ta có \(\frac{-5}{6}\)\(\times\)\(\frac{120}{25}\)< \(x\)<\(\frac{-7}{15}\)\(\times\)\(\frac{4}{9}\)\(\Rightarrow\)\(-4\)<\(x\)<\(-0,2074074074\)\(\Rightarrow\)\(-4\)<\(x\)<\(-0,2\)
mà \(x\)\(\in\)\(ℤ\)\(\Rightarrow\)\(x\)\(\in\)( -1;-2;-3)
b) ta có \(\left(\frac{-5}{3}\right)^3\)<\(x\)<\(\frac{-25}{35}\)\(\times\)\(\frac{-5}{6}\)\(\Rightarrow\)\(-4,62962963\)<\(x\)<\(0,5952380952\)
mà \(x\)\(\in\)\(ℤ\)\(\Rightarrow\)\(x\)\(\in\)(-4;-3;-2;-1;0)
ĐÚNG THÌ K CHO MK NHA
Câu b: Đặt \(B=\left(\frac{1}{2}-1\right)\cdot\left(\frac{1}{3}-1\right)\cdot\left(\frac{1}{4}-1\right)\cdot...\cdot\left(\frac{1}{2004}-1\right)\)
Ta có: \(\frac{1}{2}-1=\left(-\frac{1}{2}\right);\frac{1}{3}-1=\left(-\frac{2}{3}\right);...;\frac{1}{2004}-1=\left(-\frac{2003}{2004}\right)\)
\(\Rightarrow B=\left(-\frac{1}{2}\right)\cdot\left(-\frac{2}{3}\right)\cdot...\cdot\left(-\frac{2003}{2004}\right)\)
Vì B là 2003 thừa số âm nhân lại với nhau nên B là số âm
\(\Rightarrow B=-\left(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{2003}{2004}\right)=-\frac{1}{2004}\)
Câu a: Đặt \(A=1+2^4+2^8;B=1+2+2^2+...+2^{11}\)
\(\Rightarrow16A=2^4+2^8+2^{12}\) \(\Rightarrow15A=2^{12}-1\) \(\Rightarrow A=\frac{2^{12}-1}{15}\) \(\left(1\right)\)
\(\Rightarrow2B=2+2^2+2^3+...+2^{12}\) \(\Rightarrow B=2^{12}-1\) \(\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\) \(\Rightarrow A:B=\frac{2^{12}-1}{15}:\left(2^{12}-1\right)=\frac{1}{15}\)
Tớ biết làm đúng 100%:
\((x\cdot1+x\cdot\frac{7}{9})\left(x\cdot1+x\cdot\frac{7}{20}\right)...\left(x\cdot1+x\cdot\frac{7}{9200}\right)=\frac{186}{25}\)
\(x\cdot\left(1+\frac{7}{9}\right)\cdot x\left(1+\frac{7}{20}\right)\cdot...\cdot x\left(1+\frac{7}{9200}\right)=\frac{186}{25}\)
\(\left(x\cdot x\cdot...\cdot x\right)(\frac{16}{9}+\frac{27}{20}+...+\frac{9207}{9200})=\frac{186}{25}\)
\(\left(x\cdot x\cdot...\cdot x\right)\left(\frac{2\cdot8}{1\cdot9}+\frac{3\cdot9}{2\cdot10}+...+\frac{93\cdot99}{92\cdot100}\right)=\frac{186}{25}\)
\(x^{92}\cdot\frac{2\cdot8\cdot3\cdot9\cdot...\cdot93\cdot99}{1\cdot9\cdot2\cdot10\cdot...\cdot92\cdot100}=\frac{186}{25}\)
\(x^{92}\cdot\frac{\left(2\cdot3\cdot...\cdot93\right)\cdot\left(8\cdot9\cdot...\cdot99\right)}{\left(1\cdot2\cdot...\cdot92\right)\cdot\left(9\cdot10\cdot...\cdot100\right)}=\frac{186}{25}\)
\(x^{92}\cdot\frac{93\cdot8}{100}=\frac{186}{25}\)
\(x^{92}\cdot\frac{186}{25}=\frac{186}{25}\)
\(x^{92}=\frac{186}{25}:\frac{186}{25}\)
\(x^{92}=1\Rightarrow x=1\)
cô tớ giải rồi . x=1 (đúng 100%)
a) Ta có:
\(x-\left\{\left[-x-\left(x+3\right)\right]-\left[\left(x+2018\right)-\left(x+2019\right)\right]+21\right\}\)
\(=x-\left\{\left[-x-x-3\right]-\left[x+2018-x-2019\right]+21\right\}\)
\(=x-\left\{\left[-2x-3\right]-\left[2018-2019\right]+21\right\}\)
\(=x+2x+-3+1-21\)
\(=3x-23\)
=> \(3x-23=2020\)
\(3x=2020+23=2043\)
=> \(x=2043:3=681\)
Nhầm
\(=x-\left\{-2x-3+1+21\right\}\\ =x+2x+3-1-21\)
\(=3x-17\\ =>3x-17=2020\\ 3x=2020+17=2037\\ x=2037:3=679\)
a)
Gọi d=(2n+1;3n+2)
Ta có
2n+1\(⋮\)d => 3(2n+1)=6n+3\(⋮\)d
3n+2\(⋮\)d => 2(3n+2)=6n+4\(⋮\)d
=> 6n+4-(6n+3)=1\(⋮\)d
hay d=1
Vậy 2n+1 và 3n+2 là số nguyên tố cùng nhau
a) Gọi \(\left(2n+1;3n+2\right)=d\)
\(\Rightarrow\hept{\begin{cases}2n+1⋮d\\3n+2⋮d\end{cases}\Rightarrow\hept{\begin{cases}6n+3⋮d\\6n+4⋮d\end{cases}}}\)
\(\Rightarrow\left(6n+4\right)-\left(6n+3\right)⋮d\)
\(\Rightarrow1⋮d\)
\(\Rightarrow d\inƯ\left(1\right)=\left\{\pm1\right\}\)
Vậy 2n+1 và 3n+2 nguyên tố cùng nhau
Phần B là 20129 . ( x - 612 ) = 201210 nha
(2x-25):13=13
2x-15=169
x=92