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1.
a,
\(\left(\dfrac{2}{11\cdot13}+\dfrac{2}{13\cdot15}+...+\dfrac{2}{19\cdot21}\right)\cdot462-\left[2,04:\left(x+1,05\right)\right]:0,12=19\\ \left(\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+...+\dfrac{1}{19}-\dfrac{1}{21}\right)\cdot462-\left[2,04:\left(x+1,05\right)\right]:0,12=19\\ \left(\dfrac{1}{11}-\dfrac{1}{21}\right)\cdot462-\left[2,04:\left(x+1,05\right)\right]:0,12=19\\ \dfrac{10}{231}\cdot462-\left[2,04:\left(x+1,05\right)\right]:0,12=19\\ 20-\left[2,04:\left(x+1,05\right)\right]:0,12=19\\ \left[2,04:\left(x+1,05\right)\right]:0,12=1\\ 2,04:\left(x+1,05\right)=0,12\\ x+1,05=17\\ x=15,95\)
b,
\(\dfrac{1}{24\cdot25}+\dfrac{1}{25\cdot26}+...+\dfrac{1}{29\cdot30}+x:\dfrac{1}{3}=-4\\ \dfrac{1}{24}-\dfrac{1}{25}+\dfrac{1}{25}-\dfrac{1}{26}+...+\dfrac{1}{29}-\dfrac{1}{30}+x\cdot3=-4\\ \dfrac{1}{24}-\dfrac{1}{30}+x\cdot3=-4\\ \dfrac{1}{120}+x\cdot3=-4\\ 3x=\dfrac{-481}{120}\\ x=\dfrac{-481}{360}\)
2.
a,
\(\dfrac{15}{28}-\dfrac{186}{1116}-\dfrac{121}{462}+\dfrac{189}{198}\\ =\dfrac{15}{28}-\dfrac{1}{6}-\dfrac{11}{42}+\dfrac{21}{22}\\ =\dfrac{495}{924}-\dfrac{154}{924}-\dfrac{242}{924}+\dfrac{882}{924}\\ =\dfrac{495-154-242+882}{924}\\ =\dfrac{981}{924}\\ =\dfrac{327}{308}\)
b,
\(\left(1+\dfrac{1}{1\cdot3}\right)\cdot\left(1+\dfrac{1}{2\cdot4}\right)\cdot\left(1+\dfrac{1}{3\cdot5}\right)\cdot...\cdot\left(1+\dfrac{1}{99\cdot101}\right)\\ =\left(\dfrac{1\cdot3}{1\cdot3}+\dfrac{1}{1\cdot3}\right)\cdot\left(\dfrac{2\cdot4}{2\cdot4}+\dfrac{1}{2\cdot4}\right)\cdot\left(\dfrac{3\cdot5}{3\cdot5}+\dfrac{1}{3\cdot5}\right)\cdot...\cdot\left(\dfrac{99\cdot101}{99\cdot101}+\dfrac{1}{99\cdot101}\right)\\ =\left(\dfrac{2^2-1}{1\cdot3}+\dfrac{1}{1\cdot3}\right)\cdot\left(\dfrac{3^2-1}{2\cdot4}+\dfrac{1}{2\cdot4}\right)\cdot\left(\dfrac{4^2-1}{3\cdot5}+\dfrac{1}{3\cdot5}\right)\cdot...\cdot\left(\dfrac{100^2-1}{99\cdot101}+\dfrac{1}{99\cdot101}\right)\)\(=\dfrac{2^2}{1\cdot3}\cdot\dfrac{3^2}{2\cdot4}\cdot\dfrac{4^2}{3\cdot5}\cdot...\cdot\dfrac{100^2}{99\cdot101}\\ =\dfrac{2\cdot2}{1\cdot3}\cdot\dfrac{3\cdot3}{2\cdot4}\cdot\dfrac{4\cdot4}{3\cdot5}\cdot...\cdot\dfrac{100\cdot100}{99\cdot101}\\ =\dfrac{2\cdot2\cdot3\cdot3\cdot4\cdot4\cdot...\cdot100\cdot100}{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot...\cdot99\cdot101}\\ =\dfrac{\left(2\cdot3\cdot4\cdot...\cdot100\right)\cdot\left(2\cdot3\cdot4\cdot...\cdot100\right)}{\left(1\cdot2\cdot3\cdot...\cdot99\right)\cdot\left(3\cdot4\cdot5\cdot...\cdot101\right)}\\ =\dfrac{100\cdot2}{1\cdot101}\\ =\dfrac{200}{101}\)
5a.
\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+....+\dfrac{1}{19.21}\\ =\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+....+\dfrac{1}{19}-\dfrac{1}{21}\right)\\ =\dfrac{1}{2}\left(1-\dfrac{1}{21}\right)\\ =\dfrac{1}{2}.\dfrac{20}{21}=\dfrac{10}{21}\)
b.
\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}\\ =\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+....+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\\ =\dfrac{1}{2}\left(1-\dfrac{1}{2n+1}\right)< \dfrac{1}{2}.1=\dfrac{1}{2}\)
a, Đề sai hả bạn ??
b, \(\dfrac{\left(1,16-x\right).5,25}{\left(10\dfrac{5}{9}-7\dfrac{1}{4}\right).2\dfrac{2}{17}}=75\%\)
\(\dfrac{\left(1,16-x\right).5,25}{\left(\dfrac{95}{9}-\dfrac{29}{4}\right).\dfrac{36}{17}}=\dfrac{75}{100}\)
\(\dfrac{\left(1,16-x\right).5,25}{\left(\dfrac{380}{36}-\dfrac{261}{36}\right).\dfrac{36}{17}}=\dfrac{3}{4}\)
\(\dfrac{\left(1,16-x\right).5,25}{\dfrac{119}{36}.\dfrac{36}{17}}=\dfrac{3}{4}\)
\(\dfrac{\left(1,16-x\right).5,25}{7}=\dfrac{3}{4}\)
=> \(\left[\left(1,16-x\right).5,25\right].4=3.7\)
\(\left[\left(1,16-x\right).5,25\right].4=21\)
( 1,16 - x ) . 5,25 = 21/4
1,16 - x = 21/4 : 5,25
1,16 - x = 1
x = 1,16 - 1
x = 0,16
Vậy x = 0,16
c, \(\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{19.21}\right).420-\left[0,4.\left(7,5-2,5x\right)\right]:0,25=212\)
\(\dfrac{1}{2}.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{19.21}\right).420-\left[0,4.\left(7,5-2,5x\right)\right]:0,25=212\)
\(\dfrac{1}{2}.\left(\dfrac{1}{1}-\dfrac{1}{21}\right).420-\left[0,4.\left(7,5-2,5x\right)\right]:0,25=212\)
\(\dfrac{1}{2}.\dfrac{20}{21}.420-\left[0,4.\left(7,5-2,5x\right)\right]:0,25=212\)
\(200-\left[0,4.\left(7,5-2,5x\right)\right]:0,25=212\)
\(0,4.\left(7,5-2,5x\right):0,25=200-212\)
\(0,4.\left(7,5-2,5x\right):0,25=-12\)
0,4 . ( 7,5 - 2,5x ) = -12 . 0,25
0,4 . ( 7,5 - 2,5x ) = -3
7,5 - 2,5x = -3 :0,4
7,5 - 2,5x = -7,5
2,5x = 7,5-(-7,5)
2,5x = 15
x = 6
Vậy x = 6
Vậy x = 51
câu a chắc mk nhìn ko rõ vì mk cận mà ko đeo kính, ghi sai đề
2: Ta có: \(A=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)
\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
1: Ta có: \(A=\left(\dfrac{x-5\sqrt{x}}{x-25}-1\right):\left(\dfrac{25-x}{x+2\sqrt{x}-15}-\dfrac{\sqrt{x}+3}{\sqrt{x}+5}-\dfrac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)
\(=\left(\dfrac{x-5\sqrt{x}-x+25}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\right):\dfrac{25-x-x+9-x+25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-5}{\sqrt{x}+5}\cdot\dfrac{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}{-3x+59}\)
\(=\dfrac{-5\left(\sqrt{x}-3\right)}{-3x+59}\)
\(=\dfrac{5\sqrt{x}-15}{3x-59}\)
2: Ta có: \(A=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)
\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
a: \(A=\dfrac{-7}{28}\cdot\dfrac{15}{25}=\dfrac{-1}{4}\cdot\dfrac{3}{5}=\dfrac{-3}{20}\)
b: \(B=\dfrac{-5\cdot7}{14\cdot\left(-3\right)}=\dfrac{35}{42}=\dfrac{5}{6}\)
c: \(C=\dfrac{-1}{5}-\dfrac{1}{5}\cdot\dfrac{3}{5}=\dfrac{-1}{5}-\dfrac{3}{25}=\dfrac{-8}{25}\)
d: \(D=\dfrac{-3}{4}-\dfrac{1}{4}=-1\)
e: \(E=\dfrac{-4}{5}\left(1-\dfrac{15}{16}\right)=\dfrac{-4}{5}\cdot\dfrac{1}{16}=\dfrac{-1}{20}\)
f: \(F=\dfrac{6-7}{4}\cdot\dfrac{4+12}{22}=\dfrac{-1}{4}\cdot\dfrac{8}{11}=\dfrac{-2}{11}\)
b) ĐKXĐ : \(x\ne\pm1\)
\(P=\dfrac{x}{x-1}+\dfrac{3}{x+1}-\dfrac{6x-4}{x^2-1}\)
\(=\dfrac{x\left(x+1\right)+3\left(x-1\right)-\left(6x-4\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^2-2x+1}{\left(x-1\right)\left(x+1\right)}=\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{x-1}{x+1}\)
c) ĐKXĐ : \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
\(A=\dfrac{1}{x+\sqrt{x}}+\dfrac{2\sqrt{x}}{x-1}-\dfrac{1}{x-\sqrt{x}}\)
\(=\dfrac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}+\dfrac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}-1+2x-\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{2\left(x-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{2\left(x-1\right)}{\sqrt{x}\left(x-1\right)}=\dfrac{2}{\sqrt{x}}\)
a) ĐKXĐ : \(x\ge0;x\ne16\)
\(B=\left(\dfrac{\sqrt{x}}{\sqrt{x}+4}+\dfrac{4}{\sqrt{x-4}}\right):\dfrac{x+16}{\sqrt{x}+2}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-4\right)+4\left(\sqrt{x}+4\right)}{x-16}:\dfrac{x+16}{\sqrt{x}+2}\)
\(=\dfrac{x+16}{x-16}:\dfrac{x+16}{\sqrt{x}+2}=\dfrac{\sqrt{x}+2}{x-16}\)
a, đk x khác 0
<=> x^2 = 16 <=> x = 4 ; x = -4 (tm)
b, <=> 36x +252 = -360 <=> x = -17
c. đk x khác -1
<=> (x+1)^2 = 16
TH1 : x + 1 = 4 <=> x = 3 (tm)
TH2 : x + 1 = -4 <=> x = -5 (tm)
d, đk x khác 1/2
<=> (2x-1)^2 = 81
TH1 : 2x - 1 = 9 <=> x = 5 (tm)
TH2 : 2x - 1 = -9 <=> x = -4 (tm)
a: \(\Leftrightarrow x^2=16\)
hay \(x\in\left\{4;-4\right\}\)
b: =>x+7/15=-2/3
=>x+7=-10
hay x=-17
c: \(\Leftrightarrow\left(x+1\right)^2=16\)
\(\Leftrightarrow x+1\in\left\{4;-4\right\}\)
hay \(x\in\left\{3;-5\right\}\)
\(A=\dfrac{3}{4}.\dfrac{8}{9}.\dfrac{15}{16}.\dfrac{24}{25}.....\dfrac{99}{100}\)
\(\Leftrightarrow A=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}.\dfrac{4.6}{5.5}.....\dfrac{9.11}{10.10}\)
\(=\dfrac{1.3.2.4.3.5.4.6....9.11}{2.2.3.3.4.4.5.5.....10.10}\)
\(=\dfrac{\left(1.2.3.4.5....9\right).\left(2.3.4.5.6.....11\right)}{\left(2.3.4.5.6.....10\right)\left(2.3.4.5.6.....10\right)}\)
\(=\dfrac{11}{10}\)
còn phần a thì sao bn?