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\(\frac{|x-2|}{12}\)\(+\)\(\frac{|x-2|}{20}+\)\(\frac{|x-2|}{30}+\)\(\frac{|x-2|}{42}\)\(=\frac{70^5}{2^3.21^6}\)
\(\Rightarrow|x-2|.\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)=\frac{2^5.5^5.7^5}{2^3.7^6.3^6}\)
\(\Rightarrow|x-2|.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)=\frac{2^2.5^5}{7.3^6}\)
\(\Rightarrow|x-2|.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)=\frac{4.5^5}{21.3^5}\)
\(\Rightarrow|x-2|\left(\frac{1}{3}-\frac{1}{7}\right)=\frac{4.5^5}{21.3^5}\)\(\Rightarrow|x-2|=\frac{5^5}{3^5}\)
ĐẾN ĐÂY DỄ RÙI TỰ GIẢI TIẾP
\(\left(x-2\right):\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\right)=\frac{16}{9}\)
\(\left(x-2\right):\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)=\frac{16}{9}\)
\(\left(x-2\right):\frac{2}{9}=\frac{16}{9}\)
\(x-2=\frac{32}{91}\)
\(x=\frac{32}{91}+2\)
\(x=\frac{212}{91}\)
Đặt \(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+....+\frac{1}{90}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{9.10}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{9}-\frac{1}{10}\)
\(A=1-\frac{1}{10}\)
\(A=\frac{9}{10}\)
\(=>\left[y-\frac{1}{2}\right]x\frac{9}{10}=\frac{1}{3}\)
\(y-\frac{1}{2}=\frac{1}{3}:\frac{9}{10}\)
\(y-\frac{1}{2}=\frac{10}{27}\)
\(=>y=\frac{10}{27}+\frac{1}{2}\)
\(y=\frac{20+27}{54}=\frac{47}{54}\)
Vậy \(y=\frac{47}{54}\)
Ủng hộ mk nha!!!
a) \(14:\frac{0,4x+0,6}{x}=7\)
\(\frac{0,4x+0,6}{x}=2\)
0,4x + 0,6 = 2.x
2x - 0,4x = 0,6
1,6x = 0,6
x = 0,375
b) \(\left(160\%+\frac{2}{3}x-x\right).12=660\)
\(\left(160\%+\frac{2}{3}x-x\right)=55\)
\(x\left(\frac{2}{3}-1\right)=53,4\)
\(-\frac{1}{3}x=\frac{267}{5}\)
\(x=\frac{267}{5}.\frac{3}{-1}\)
\(x=-160,2\)
c) \(1:\frac{1.2.3.4.....31}{2.2.2.3.2.4.....2.32}=2^x\)
\(1:\frac{1.2.3.4.....31}{2^{31}.2.3.4.....31.2^5}=2^x\)
\(1:\frac{1}{2^{36}}=2^x\)
\(2^{36}=2^x\)
\(x=36\)
b \(\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+...+\frac{1}{x\cdot\left(x+1\right)}=\frac{19}{100}\)
=>\(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{19}{100}\)
=>\(\frac{1}{5}-\frac{1}{x+1}\)\(=\frac{19}{100}\)
=>\(\frac{1}{x+1}=\frac{1}{5}-\frac{19}{100}\)
=>\(\frac{1}{x+1}=\frac{1}{100}\)
=> x+1 =100
=>x=99
b) \(\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{x\left(x+1\right)}=\frac{19}{100}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{19}{100}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{x+1}=\frac{19}{100}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{5}-\frac{19}{100}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{100}\)
\(\Rightarrow x+1=100\)
\(\Rightarrow x=99\)
c) \(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{x\left(x+2\right)}=\frac{49}{99}\)
\(\Rightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{49}{99}\)
\(\Rightarrow1-\frac{1}{x+2}=\frac{49}{99}\)
\(\Rightarrow\frac{1}{x+2}=1-\frac{49}{99}\)
\(\Rightarrow\frac{1}{x+2}=\frac{50}{99}\)
\(\Rightarrow50.\left(x+2\right)=99\)
\(\Rightarrow x+2=\frac{99}{50}\)
\(\Rightarrow x=-\frac{1}{99}\)
d) Ta có : 6 = 1.6 = 2.3 = (-2) . (-3)
Lâp bảng xét 6 trường hợp:
| \(2x+1\) | \(1\) | \(6\) | \(2\) | \(3\) | \(-2\) | \(-3\) |
| \(y-2\) | \(6\) | \(1\) | \(3\) | \(2\) | \(-3\) | \(-2\) |
| \(x\) | \(0\) | \(\frac{5}{2}\) | \(\frac{1}{2}\) | \(1\) | \(-\frac{3}{2}\) | \(-2\) |
| \(y\) | \(8\) | \(3\) | \(5\) | \(4\) | \(-1\) | \(0\) |
Vậy các cặp (x,y) \(\inℤ\)thỏa mãn là : (0;4) ; (1; 4) ; (-2 ; 0)
e) \(x^2-3xy+3y-x=1\)
\(\Rightarrow x\left(x-3y\right)+3y-x=1\)
\(\Rightarrow x\left(x-3y\right)-\left(x-3y\right)=1\)
\(\Rightarrow\left(x-3y\right)\left(x-1\right)=1\)
Lại có : 1 = 1.1 = (-1) . (-1)
Lập bảng xét các trường hợp :
| \(x-1\) | \(1\) | \(-1\) |
| \(x-3y\) | \(1\) | \(-1\) |
| \(x\) | \(2\) | \(0\) |
| \(y\) | \(\frac{1}{3}\) | \(\frac{1}{3}\) |
Vậy các cặp(x,y) thỏa mãn là : \(\left(2;\frac{1}{3}\right);\left(0;\frac{1}{3}\right)\)
Tìm x, biết:
3(x+2)(x+5) +5(x+5)(x+10) +7(x+10)(x+17) =x(x+2)(x+17) (x∉−2;−5;−10;−17)
2(x−1)(x−3) +5(x−3)(x−8) +12(x−8)(x−20) −1x−20 =−34 (x∉1;3;8;20)
x+110 +2+111 x+112 =x+113 +x+114
x−1030 +x−1443 +x−595 +x−1488 =0
\(\frac{47}{12}\left|x\right|=1\frac{11}{31}\cdot4\frac{3}{7}-\left(1,5-3\frac{1}{3}\right)\)
=> \(\frac{47}{12}\left|x\right|=\frac{42}{31}\cdot\frac{31}{7}-\left(\frac{3}{2}-\frac{10}{3}\right)\)
=> \(\frac{47}{12}\left|x\right|=6-\left(-\frac{11}{6}\right)\)
=> \(\frac{47}{12}\left|x\right|=\frac{47}{6}\)
=> \(\left|x\right|=\frac{47}{6}:\frac{47}{12}\)
=> \(\left|x\right|=\frac{47}{6}\cdot\frac{12}{47}\)
=> \(\left|x\right|=2\)
=> \(x\in\left\{2;-2\right\}\)
\(x+\left(\frac{-31}{12}\right)^2\left(\frac{49}{12}\right)^2-x=y\)
\(x+\frac{961}{144}.\frac{2401}{144}-x=y\)
\(x+\frac{2307361}{20736}-x=y\)
\(y=\frac{2307361}{20736}\)
Thay vào \(x+\frac{961}{144}.\frac{2401}{144}-x=y\) ta được
\(x+\frac{2307361}{20736}-x=y\)
\(x-x+\frac{2307361}{20736}=\frac{2307361}{20736}\)
Vậy x thuộc N;\(y=\frac{2307361}{20736}\)
\(\Leftrightarrow\left(x-x\right)+\frac{961}{144}\cdot\frac{2401}{144}=y\Leftrightarrow y=\frac{2307361}{20736}\)
\(\Rightarrow x\in R\)
( hình như đề sai :v )