Bài 1: Tính: A= \(\frac{81^{10}.3^{17}}{27^{10}.9^{13}}\)
Bài 2: Tìm x, y: \(\frac{x}{7}=-\frac{12}{49}\)
bài 1: cho x, y thuộc Q. cmr:
|x + y| =< |x| + |y|
bài 2: tính:
\(A=\frac{\left(13\frac{1}{4}-2\frac{5}{27}-10\frac{5}{6}\right).230\frac{1}{25}+46\frac{3}{4}}{\left(1\frac{3}{7}+\frac{10}{3}\right):\left(12\frac{1}{3}-14\frac{2}{7}\right)}\)
bài 3: cho a + b + c = a^2 + b^2 + c^2 = 1 và x : y : z = a : b : c.
cmr: (x + y + z)^2 = x^2 + y^2 + z^2
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Bài 1:
Với mọi gt \(x,y\in Q\) ta luôn có:
\(x\le\left|x\right|\) và \(-x\le\left|x\right|\)
\(y\le\left|y\right|\) và \(-y\le\left|y\right|\Rightarrow x+y\le\left|x\right|+\left|y\right|\) và \(-x-y\le\left|x\right|+\left|y\right|\)
Hay: \(x+y\ge-\left(\left|x\right|+\left|y\right|\right)\)
Do đó: \(-\left(\left|x\right|+\left|y\right|\right)\le x+y\le\left|x\right|+\left|y\right|\)
Vậy: \(\left|x+y\right|\le\left|x\right|+\left|y\right|\)
Dấu "=" xảy ra khi: \(xy\ge0\)
Bài 3:
Ta có: \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=\frac{x+y+z}{a+b+c}=x+y+z\) (vì a + b + c = 1)
Do đó: \(\left(x+y+z\right)^2=\frac{x^2}{a^2}=\frac{y^2}{b^2}=\frac{z^2}{c^2}=\frac{x^2+y^2+z^2}{a^2+b^2+c^2}=x^2+y^2+z^2\) (vì a2 + b2 + c2 = 1)
Vậy: (x + y + z)2 = x2 + y2 + z2
bài 1 : tính hợp lí
a) \( \left(-0,125\right)^3.80^4\)
b) \(\frac{81^{11}.3^{17}}{27^{10}.9^{15}}\)
bài 2 : rút gọn
a) \(A=\frac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}\)
b)\(B=\left(6^9.2^{10}+12^{10}\right):\left(2^{19}.27^3+15.4^9.9^4\right)\)
c)\(H=x+\frac{0,2-0,375+\frac{5}{11}}{-0,3+\frac{9}{16}-\frac{15}{22}}vs,x=-\frac{1}{3}\)
\(A=\frac{\frac{1}{11}-\frac{1}{13}-\frac{1}{17}}{\frac{5}{11}-\frac{5}{13}-\frac{5}{17}}+\frac{\frac{-3}{3}-\frac{2}{9}-\frac{2}{27}+\frac{2}{81}}{\frac{7}{3}-\frac{7}{9}-\frac{7}{27}+\frac{7}{81}}\)
A = \(\frac{\frac{1}{11}-\frac{1}{13}-\frac{1}{17}}{\frac{5}{11}-\frac{5}{13}-\frac{5}{17}}+\frac{\frac{2}{3}-\frac{2}{9}-\frac{2}{27}+\frac{2}{81}}{\frac{7}{3}-\frac{7}{9}-\frac{7}{27}+\frac{7}{81}}\)
B = \(\frac{5^2}{11.16}+\frac{5^2}{16.21}+\frac{5^2}{21.26}+\frac{5^2}{26.31}+...+\frac{5^2}{56.81}\)
C = \(1-\frac{1}{3}-\frac{1}{6}-\frac{1}{10}-\frac{1}{15}-...-\frac{1}{1225}\)
a)
\(\Rightarrow A=\frac{\frac{1}{11}-\frac{1}{13}-\frac{1}{17}}{5\left(\frac{1}{11}-\frac{1}{13}-\frac{1}{17}\right)}+\frac{2\left(\frac{1}{3}-\frac{1}{9}-\frac{1}{27}+\frac{1}{81}\right)}{7\left(\frac{1}{3}-\frac{1}{9}-\frac{1}{27}+\frac{1}{81}\right)}\)
\(\Rightarrow A=\frac{1}{5}+\frac{2}{7}\)
\(\Rightarrow A=\frac{17}{35}\)
b)
\(\Rightarrow B=5\left(\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+....+\frac{1}{56}-\frac{1}{61}\right)\)
\(\Rightarrow B=5\left(\frac{1}{11}-\frac{1}{61}\right)\)
\(\Rightarrow B=5.\frac{50}{671}=\frac{250}{671}\)
c)
\(\Rightarrow C=1-\left(\frac{1}{1.3}+\frac{1}{2.3}+\frac{1}{2.5}+....+\frac{1}{49.25}\right)\)
\(\Rightarrow C=1-2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{49.50}\right)\)
\(\Rightarrow C=1-2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{49}-\frac{1}{50}\right)\)
\(\Rightarrow C=1-1-\frac{1}{25}\)
\(\Rightarrow C=\frac{1}{25}\)
Bài 1: Thực hiện các phép tính dau bằng cách hợp lí
a. \(\frac{11}{225}-\frac{17}{18}-\frac{5}{7}+\frac{4}{9}+\frac{17}{14}\)
b. \(1-\frac{1}{2}+2-\frac{2}{3}+3-\frac{3}{4}+4-\frac{1}{4}-3-\frac{1}{3}-2-\frac{1}{2}-1\)
Bài 2: Tìm x biết
a. \(\frac{11}{13}-\left(\frac{5}{42}-x\right)=-\left(\frac{15}{28}-\frac{11}{13}\right)\)
b. \(\left|x+\frac{4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\)
Bài 3: Thực hiện các phép tính sau bằng cách hợp lí nhất
a. \(\left(-\frac{40}{51}\cdot0,32\cdot\frac{17}{20}\right):\frac{64}{75}\)
b. \(-\frac{10}{11}\cdot\frac{8}{9}+\frac{7}{18}\cdot\frac{10}{11}\)
c. \(\frac{3}{14}:\frac{1}{28}-\frac{13}{21}:\frac{1}{28}+\frac{29}{42}-8\)
d. \(-1\frac{5}{7}\cdot15+\frac{2}{7}.\left(-15\right)+\left(-105\right).\left(\frac{2}{3}-\frac{4}{5}+\frac{1}{7}\right)\)
Bìa 4: Tính giá trị của các biểu thức sau
a. \(A=7x-2x-\frac{2}{3}y+\frac{7}{9}y\) với \(x=-\frac{1}{10};y=4,8\)
b. \(B=x+\frac{0,2-0,375+\frac{5}{11}}{-0,3+\frac{9}{16}-\frac{15}{22}}\) với\(x=-\frac{1}{3}\)
a) Tìm số nguyên x, biết:
\(\frac{x}{9}=\frac{-12}{27}\)
b) Tìm số nguyên x, biết: 12 - ( x - 4 ) = 17
c) Tìm y biết: \(\left(1\frac{2}{3}+2\frac{2}{3}y\right).\frac{10}{11}=2\frac{3}{11}\)
a) Ta có: \(\frac{x}{9}=\frac{-12}{27}\)
=> \(27.x=-12.9\)
=> \(27x=-108\)
=> \(x=108:27\)
=>\(x=4\)
bài 1: Tìm x,y biết rằng:
\(x+(-\frac{31}{12})^2=\left(\frac{49}{12}\right)^2-x=y^2\)
bài 2: tìm x biết:
a.\(5^x.\left(5^3\right)^2=625\) b.\(\left(\frac{12}{25}\right)^x=\left(\frac{5}{3}\right)^{-2}-\left(-\frac{3}{5}\right)^4\) c.\(\left(-\frac{3}{4}\right)^{3x-1}=\frac{256}{81}\)
d.\(172x^2-7^9:98^3=2^{-3}\)
Bài 3: Tìm x \(\varepsilon\)N biết:
a.\(8< 2^x\le2^9\times2^{-5}\) b.\(27< 81^3:3^x< 243\) \(\left(\frac{2}{5}\right)^x>\left(\frac{5}{2}\right)^{-3}\times\left(-\frac{2}{5}\right)^2\)c.
Bài 1:
Ta có: \(x+\left(-\frac{31}{12}\right)^2=\left(\frac{49}{12}\right)^2-x\)
\(\Leftrightarrow2x=\frac{1440}{144}=10\)
\(\Rightarrow x=5\)
Khi đó: \(y^2=\left(\frac{49}{12}\right)^2-5=\frac{1681}{144}\)
=> \(\hept{\begin{cases}y=\frac{41}{12}\\y=-\frac{41}{12}\end{cases}}\)
Tính :
a ) \(A=\frac{\frac{1}{11}-\frac{1}{13}-\frac{1}{17}}{\frac{5}{11}-\frac{5}{13}-\frac{5}{17}}+\frac{\frac{2}{3}-\frac{2}{9}-\frac{2}{27}+\frac{2}{81}}{\frac{7}{3}-\frac{7}{9}-\frac{7}{27}+\frac{7}{81}}\)
b ) \(B=\frac{5^2}{11.16}+\frac{5^2}{16.21}+\frac{5^2}{21.26}+\frac{5^2}{26.31}+...+\frac{5^2}{56.61}\)
c ) \(C=-1-\frac{1}{3}-\frac{1}{6}-\frac{1}{10}-\frac{1}{15}-...-\frac{1}{1225}\)
Cố giúp mk nha mai mk nộp rùi đó
a) \(A=\frac{\frac{1}{11}-\frac{1}{13}-\frac{1}{17}}{\frac{5}{11}-\frac{5}{13}-\frac{5}{17}}+\frac{\frac{2}{3}-\frac{2}{9}-\frac{2}{27}+\frac{2}{81}}{\frac{7}{3}-\frac{7}{9}-\frac{7}{27}+\frac{7}{81}}\)
\(=\frac{\frac{1}{11}-\frac{1}{13}-\frac{1}{17}}{5\left(\frac{1}{11}-\frac{1}{13}-\frac{1}{17}\right)}+\frac{2\left(\frac{1}{3}-\frac{1}{9}-\frac{1}{27}+\frac{1}{81}\right)}{7\left(\frac{1}{3}-\frac{1}{9}-\frac{1}{27}+\frac{1}{81}\right)}\)
\(=\frac{1}{5}+\frac{2}{7}\)
\(=\frac{7}{35}+\frac{10}{35}\)
\(=\frac{17}{35}\)
Vậy \(A=\frac{17}{35}\)
b) \(B=\frac{5^2}{11.16}+\frac{5^2}{16.21}+\frac{5^2}{21.26}+\frac{5^2}{26.31}+...+\frac{5^2}{56.61}\)
\(=5.\left(\frac{5}{11.16}+\frac{5}{16.21}+\frac{5}{21.26}+...+\frac{5}{56.61}\right)\)
\(=5.\left(\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+\frac{1}{21}-\frac{1}{26}+...+\frac{1}{56}-\frac{1}{61}\right)\)
\(=5.\left(\frac{1}{11}-\frac{1}{61}\right)\)
\(=5.\left(\frac{61}{671}-\frac{11}{671}\right)\)
\(=5.\frac{50}{671}\)
\(=\frac{250}{671}\)
Vậy \(B=\frac{250}{671}\)
Bài 1 : Tính
A = \(\frac{81^4.3^{10}.27^5:3^{12}}{3^{18}:9^3.243^2}\)
B = \(\frac{2.55^2-9.55^{21}}{25^{10}}:\frac{5.\left(3.7^{15}-19\cdot7^{14}\right)}{7^{16}+3.7^{15}}\)
\(A=\frac{81^4.3^{10}.27^5:3^{12}}{3^{18}:9^3.243^2}=\frac{3^{16}.3^{10}.3^{15}:3^{12}}{3^{18}:3^6.3^{10}}=\frac{3^{29}}{3^{22}}=3^7\)
\(B=\frac{2.55^2-9.55^{21}}{25^{10}}:\frac{5\left(3.7^{15}-19.7^{14}\right)}{7^{16}+3.7^{15}}=\frac{2.55^2-9.55^2.55^{19}}{25^{10}}:\frac{5\left(21.7^{14}-19.7^{14}\right)}{7.7^{15}+3.7^{15}}=\frac{55^2\left(55^{19}.9-2\right)}{25^{10}}:\frac{5.7^{14}.2}{7^{15}.10}=\frac{55^2\left(55^{19}.9-2\right)}{25^{10}}.\frac{7^{15}.10}{5.7^{14}.2}\)Chịu ==