Tính:
\(\left(\dfrac{1}{10}\right)^{10}.\left(\dfrac{1}{2}\right)^{50}\)
Trong vở bài tập của bạn Dũng có bài làm sau :
a) \(\left(-5\right)^2.\left(-5\right)^3=\left(-5\right)^6\)
b) \(\left(0,75\right)^3:0,75=\left(0,75\right)^2\)
c) \(\left(0,2\right)^{10}:\left(0,2\right)^5=\left(0,2\right)^2\)
d) \(\left[\left(-\dfrac{1}{7}\right)^2\right]^4=\left(-\dfrac{1}{7}\right)^6\)
e) \(\dfrac{50^3}{125}=\dfrac{50^3}{5^3}=\left(\dfrac{50}{5}\right)^3=10^3=1000\)
Hãy kiểm tra lại các đáp số và sửa lại chỗ sai (nếu có)
f) \(\dfrac{8^{10}}{4^8}=\left(\dfrac{8}{4}\right)^{10-8}=2^2\)
Các câu sai: a, c, d, f
Các câu đúng: b, e
Các câu đúng: b,e
Các câu sai: a, c, d; f.
a) \(\left(-5\right)^2.\left(-5\right)^3=\left(-5\right)^5\);
c) \(\left(0,2\right)^{10}:\left(0,2\right)^5=\left(0,2\right)^{10-5}=0,2^5\);
d) \(\left[\left(-\dfrac{1}{7}\right)^2\right]^4=\left(-\dfrac{1}{7}\right)^{2.4}=\left(-\dfrac{1}{7}\right)^8\)
f \(\dfrac{8^{10}}{4^8}=\dfrac{\left(2^3\right)^5}{\left(2^2\right)^8}=\dfrac{2^{15}}{2^{16}}=\dfrac{1}{2}\)
- Các câu sai là a, c, d, f
- Các câu đúng là b, e
Sửa lại:
a) (-5)2 .(-5)3 = (-5)5
c) (0,2)10 : (0,2)5 = (0,2)5
Tính
a) \(\dfrac{13}{50}.\left(-15.5\right):\dfrac{13}{50}.84\dfrac{1}{2}\)
b) \(\dfrac{\left(-0,7\right)^2.\left(-5\right)^3}{\left(-2\dfrac{1}{3}\right)^3.\left(1\dfrac{1}{2}\right)^4.\left(-1\right)^5}\)
\(a,=\dfrac{13}{50}\cdot\dfrac{50}{13}\cdot\left(-\dfrac{31}{2}\right)\cdot\dfrac{169}{2}=-\dfrac{5239}{2}\\ b,=\dfrac{-\dfrac{49}{100}\cdot\left(-125\right)}{-\dfrac{343}{27}\cdot\dfrac{81}{16}\cdot\left(-1\right)}=\dfrac{\dfrac{245}{4}}{\dfrac{1029}{16}}=\dfrac{245}{4}\cdot\dfrac{16}{1029}=\dfrac{20}{21}\)
a) \(\dfrac{13}{50}.\left(-15.5\right):\dfrac{13}{50}.84\dfrac{1}{2}=\dfrac{13}{50}.-75:\dfrac{13}{50}.\dfrac{169}{2}=-\dfrac{75.169}{2}=-\dfrac{12675}{2}\)
b) \(\dfrac{\left(-0,7\right)^2.\left(-5\right)^3}{\left(-2\dfrac{1}{3}\right)^3.\left(1\dfrac{1}{2}\right)^4.\left(-1\right)^5}=\dfrac{0,49.\left(-125\right)}{-\dfrac{343}{27}.\dfrac{81}{16}.\left(-1\right)}=-\dfrac{\dfrac{245}{4}}{\dfrac{1029}{16}}=\dfrac{20}{21}\)
Thực hiện phép tính sau:
\(\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)...\left(1-\dfrac{1}{10^2}\right)\).
Ta có: \(\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)\cdot...\cdot\left(1-\dfrac{1}{10^2}\right)\)
\(=\dfrac{-3}{4}\cdot\dfrac{-8}{9}\cdot\dfrac{-15}{16}\cdot...\cdot\dfrac{-99}{100}\)
\(=-\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot...\cdot\dfrac{99}{100}\)
\(=-\dfrac{10+1}{2\cdot10}=\dfrac{-11}{20}\)
Giải phương trình \(1+\dfrac{2}{x-2}=\dfrac{-10}{x+3}+\dfrac{50}{\left(2-x\right)\left(x+3\right)}\)
\(1+\dfrac{2}{x-2}=\dfrac{-10}{x+3}+\dfrac{50}{\left(2-x\right)\left(x+3\right)}\left(ĐK:x\ne2;x\ne-3\right)\)
\(\Leftrightarrow\dfrac{\left(2-x\right)\left(x+3\right)}{\left(2-x\right)\left(x+3\right)}-\dfrac{2}{2-x}=\dfrac{-10\left(2-x\right)}{\left(2-x\right)\left(x+3\right)}+\dfrac{50}{\left(2-x\right)\left(x+3\right)}\)
\(\Leftrightarrow2x+6-x^2-3x-2=-20+10x+50\)
\(\Leftrightarrow-x^2+2x-3x-10x+6-2+20-50=0\)
\(\Leftrightarrow-x^2-11x-26=0\)
\(\Leftrightarrow-\left(x^2+2x-13x+26\right)=0\)
\(\Leftrightarrow x\left(x+2\right)-13\left(x+2\right)=0\)
\(\Leftrightarrow\left(x-13\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-13=0\\x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=13\\x=-2\end{matrix}\right.\)
\(6\dfrac{9}{10}+\left(\dfrac{2}{5}-\dfrac{1}{10}\right).50\%\)
\(\dfrac{69}{10}+\left(\dfrac{4}{10}-\dfrac{1}{10}\right)x\dfrac{1}{2}=\dfrac{69}{10}+\dfrac{3}{10}x\dfrac{1}{2}=\dfrac{69}{10}+\dfrac{3}{20}=\dfrac{138}{20}+\dfrac{3}{20}=\dfrac{141}{20}\)
bài 6: tính :
\(\dfrac{10^9.\left(-81\right)^{10}}{\left(-8\right)^4.25^5.9^{10}}\)
b,\(\dfrac{9^4.\left(-4\right)^5.25^3}{8^3,\left(-27\right)^2.5^7}\)
c,\(\dfrac{3^{186}.\left(-25\right)^{50}}{\left(-15\right)^{100}.27^{29}}\)
a: \(=\dfrac{2^9\cdot5^9\cdot3^{40}}{2^{12}\cdot5^{10}\cdot3^{20}}=\dfrac{3^{20}}{5\cdot2^3}\)
b: \(=\dfrac{-3^8\cdot2^{10}\cdot5^6}{2^9\cdot\left(-1\right)\cdot3^6\cdot5^7}=\dfrac{-2}{5}\cdot3^2=-\dfrac{18}{5}\)
c: \(=\dfrac{3^{186}\cdot5^{100}}{5^{100}\cdot3^{187}}=\dfrac{1}{3}\)
1, Tính hợp lí
\(A=\dfrac{0.5+\dfrac{7}{12}-\dfrac{5}{6}}{1-\dfrac{2}{3}+0,75}\)
\(B=-66.\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{11}\right)+124.\left(-37\right)+126.\left(-62\right)\)
\(N=\left(60\dfrac{7}{13}+50\dfrac{8}{13}-11.\dfrac{2}{13}\right)x\) (Với \(x=-2017\dfrac{7}{10}\))
\(M=\left(1-\dfrac{2}{2.3}\right).\left(1-\dfrac{2}{3.4}\right).\left(1-\dfrac{2}{4.5}\right).....\left(1-\dfrac{2}{99.100}\right)\)
1) giải phương trình :
a) \(\left(2+3\right)\left(\dfrac{3x+8}{2-7x}+1\right)=\left(x-5\right)\left(\dfrac{3x+8}{2-7x}+1\right)\)
b) \(\dfrac{7x+10}{x+1}\left(x^2-x-2\right)-\dfrac{7x+10}{x+1}\left(2x^2-3x-5\right)=0\)
c) \(\dfrac{2x+5}{x+3}+1=\dfrac{4}{x^2+2x-3}-\dfrac{3x-1}{1-x}\)
d) \(\dfrac{13}{2x^2+x-21}+\dfrac{1}{2x+7}+\dfrac{6}{9-x^2}=0\)
i) \(\dfrac{x-49}{50}+\dfrac{x-50}{49}=\dfrac{49}{x-50}+\dfrac{50}{x-49}\)
k) \(\dfrac{1+\dfrac{x}{x+3}}{1-\dfrac{x}{x+3}}=3\)
b: \(\Leftrightarrow\dfrac{7x+10}{x+1}\left(x^2-x-2-2x^2+3x+5\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\left(-x^2+2x+3\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\left(x^2-2x-3\right)=0\)
=>(7x+10)(x-3)=0
hay \(x\in\left\{-\dfrac{10}{7};3\right\}\)
d: \(\Leftrightarrow\dfrac{13}{2x^2+7x-6x-21}+\dfrac{1}{2x+7}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\dfrac{13}{\left(2x+7\right)\left(x-3\right)}+\dfrac{1}{\left(2x+7\right)}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow26x+91+x^2-9-12x-14=0\)
\(\Leftrightarrow x^2+14x+68=0\)
hay \(x\in\varnothing\)
Giải pt
\(1+\dfrac{2}{x-2}=\dfrac{10}{x+3}-\dfrac{50}{\left(2-x\right)\left(x+3\right)}\)
\(\dfrac{x^2-3x+5}{x^2-4}=-1\)
a: \(\Leftrightarrow x^2+x-6+2x-6=10x-20+50\)
\(\Leftrightarrow x^2+3x-12-10x-30=0\)
\(\Leftrightarrow x^2-7x-42=0\)
\(\text{Δ}=\left(-7\right)^2-4\cdot1\cdot\left(-42\right)=217>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{7-\sqrt{217}}{2}\\x_2=\dfrac{7+\sqrt{217}}{2}\end{matrix}\right.\)
b: \(\Leftrightarrow x^2-3x+5=-x^2+4\)
\(\Leftrightarrow2x^2-3x+1=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x-1\right)=0\)
hay \(x\in\left\{\dfrac{1}{2};1\right\}\)