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Đặt \(S=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2015}-\dfrac{1}{2016}\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2016}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}\right)-\left(1+\dfrac{1}{2}+...+\dfrac{1}{1008}\right)\)
\(=\dfrac{1}{1009}+\dfrac{1}{1010}+...+\dfrac{1}{2015}+\dfrac{1}{2016}\)
Nên:
\(A=\left(\dfrac{1}{1009}+\dfrac{1}{1010}+...+\dfrac{1}{2015}+\dfrac{1}{2016}\right):\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2015}-\dfrac{1}{2016}\right)\)\(=\left(\dfrac{1}{1009}+\dfrac{1}{1010}+...+\dfrac{1}{2015}+\dfrac{1}{2016}\right):\left(\dfrac{1}{1009}+\dfrac{1}{1010}+...+\dfrac{1}{2015}+\dfrac{1}{2016}\right)\)\(\Rightarrow A=1\)
Vậy A = 1
Chúc bạn học tốt!!
\(\left(x-2\right)\left(y+3\right)=5\)
\(\Rightarrow x-2;y+3\inƯ\left(5\right)\)
\(Ư\left(5\right)=\left\{\pm1;\pm5\right\}\)
Xét ước
\(xy-6x-3y=7\)
\(\Rightarrow xy-6x-3y+18=25\)
\(\Rightarrow x\left(y-6\right)-3\left(y-6\right)=25\)
\(\Rightarrow\left(x-3\right)\left(y-6\right)=25\)
Xét ước
\(\dfrac{a}{2}-\dfrac{1}{b}=\dfrac{3}{4}\)
\(\Rightarrow\dfrac{1}{b}=\dfrac{3}{4}+\dfrac{a}{2}\)
\(\Rightarrow\dfrac{1}{b}=\dfrac{3}{4}+\dfrac{2a}{4}\)
\(\Rightarrow\dfrac{1}{b}=\dfrac{3+2a}{4}\)
\(\Rightarrow b\left(3+2a\right)=4\)
Xét ước
Bài 2: a) \(\dfrac{x-3}{x+5}=\dfrac{5}{7}\)
\(\Leftrightarrow\left(x-3\right).7=\left(x+5\right).5\)
\(\Leftrightarrow7x-21=5x+25\)
\(\Leftrightarrow7x-5x=21+25\)
\(\Leftrightarrow2x=46\)
\(\Rightarrow x=46:2=23\)
b) \(\dfrac{7}{x-1}=\dfrac{x+1}{9}\)
\(\Leftrightarrow\left(x+1\right)\left(x-1\right)=63\)
\(\Leftrightarrow x^2-1=63\)
\(\Leftrightarrow x^2=64\)
\(\Rightarrow x^2=\left(\pm8\right)^2\)
\(\Rightarrow x=8\) hoặc \(x=-8\)
2)a) \(\dfrac{x-3}{x+5}=\dfrac{5}{7}\)
\(\Leftrightarrow7\left(x-3\right)=5\left(x+5\right)\)
\(7x-21=5x+25\)
\(7x-5x+25=21\)
\(2x+25=21\)
\(2x=-4\Rightarrow x=-2\)
b) \(\dfrac{7}{x-1}=\dfrac{x+1}{9}\)
\(7.9=\left(x+1\right)\left(x-1\right)\)
\(63=x\left(x-1\right)+1\left(x-1\right)\)
\(63=x^2-x+x-1\)
\(x^2=63+1=64\)
\(x=\left\{\pm8\right\}\)
c) \(\dfrac{x+4}{20}=\dfrac{2}{x+4}\)
\(\Leftrightarrow\left(x+4\right)\left(x+4\right)=2.20=40\)
\(x\left(x+4\right)+4\left(x+4\right)=40\)
\(x^2+4x+4x+16=40\)
\(x^2+8x=40-16=24\)
\(x\left(x+8\right)=24\)
\(x\in\left\{\varnothing\right\}\)
d) \(\dfrac{x-1}{x+2}=\dfrac{x-2}{x+3}\)
\(\Leftrightarrow\left(x+2\right)\left(x-2\right)=\left(x-1\right)\left(x+3\right)\)
\(x\left(x-2\right)+2\left(x-2\right)=x\left(x+3\right)-1\left(x+3\right)\)
\(x^2-2x+2x-4=x^2+3x-x-3\)
\(\)\(x^2-4=x^2+2x-3\)
\(\Leftrightarrow x^2-x^2-2x+3=4\)
\(-2x+3=4\)
\(-2x=1\)
\(x=-\dfrac{1}{2}\)
bài 3:
a, đặt \(\dfrac{x}{12}=\dfrac{y}{9}=\dfrac{z}{5}=k\)
=>x=12k,y=9k,z=5k
ta có: ayz=20=> 12k.9k.5k=20
=> (12.9.5)k^3=20
=>540.k^3=20
=>k^3=20/540=1/27
=>k=1/3
=>x=12.1/3=4
y=9.1/3=3
z=5.1/3=5/3
vậy x=4,y=3,z=5/3
b,ta có: \(\dfrac{x}{5}=\dfrac{y}{7}=\dfrac{z}{3}=\dfrac{x^2}{25}=\dfrac{y^2}{49}=\dfrac{z^2}{9}\)
A/D tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{5}=\dfrac{y}{7}=\dfrac{z}{3}=\dfrac{x^2}{25}=\dfrac{y^2}{49}=\dfrac{z^2}{9}=\dfrac{x^2+y^2-z^2}{25+49-9}=\dfrac{585}{65}=9\)
=>x=5.9=45
y=7.9=63
z=3*9=27
vậy x=45,y=63,z=27
Không đúng
theo mk nghĩ là bài này áp dụng dãy tỉ số = nhau
a) Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{4}=\dfrac{y}{3}=\dfrac{z}{9}=\dfrac{x-3y+4z}{4-3.3+4.9}=\dfrac{63}{31}=2\)
\(\Rightarrow x=8\)
\(\Rightarrow y=6\)
\(\Rightarrow z=18\)
b. c. Xem lại đề.
2)\(x+y+z=9^2=81\)
Ta có:\(\dfrac{x}{3}=\dfrac{y}{4}\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}\left(1\right)\)
\(\dfrac{y}{5}=\dfrac{z}{7}\Rightarrow\dfrac{y}{20}=\dfrac{z}{28}\left(2\right)\)
Từ (1) và (2)\(\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}\)
\(\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}=\dfrac{x+y+z}{15+20+28}=\dfrac{81}{63}=\dfrac{9}{7}\)
\(\Rightarrow x=\dfrac{135}{7};y=\dfrac{180}{7};z=36\)
a) \(\dfrac{5x-3}{3-2x}=\dfrac{2}{3}\)
\(\Rightarrow3\left(5x-3\right)=2\left(3-2x\right)\)
\(\Rightarrow15x-9=6-4x\)
\(\Rightarrow15x+4x=9+6\)
\(\Rightarrow19x=15\Rightarrow x=\dfrac{15}{19}\)
b) \(\left(\dfrac{4}{5}x+\dfrac{2}{3}\right):\dfrac{3}{4}=2\)
\(\Rightarrow\dfrac{4}{5}x+\dfrac{2}{3}=\dfrac{3}{2}\Rightarrow\dfrac{4}{5}x=\dfrac{5}{6}\)
\(\Rightarrow x=\dfrac{25}{24}\)
c) \(\dfrac{3}{4}x-\dfrac{1}{3}=\dfrac{3}{5}\Rightarrow\dfrac{3}{4}x=\dfrac{14}{15}\)
\(\Rightarrow x=\dfrac{56}{45}\)
d) \(\dfrac{2}{3}-\dfrac{3}{5}:x=\dfrac{1}{4}\Rightarrow\dfrac{3}{5}:x=\dfrac{5}{12}\)
\(\Rightarrow x=\dfrac{36}{25}\)
Lời giải:
Từ \(\frac{x}{y}=\frac{2}{3}\Rightarrow x=\frac{2}{3}y\). Thay vào biểu thức $M$ ta có:
\(M=\frac{5x+3y}{6x-7y}=\frac{5.\frac{2}{3}y+3y}{6.\frac{2}{3}y-7y}=\frac{y(\frac{10}{3}+3)}{y(4-7)}=\frac{-19}{9}\)
Cách khác ạ :
\(\dfrac{x}{y}=\dfrac{2}{3}\Leftrightarrow\dfrac{x}{2}=\dfrac{y}{3}\)
Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=k\Leftrightarrow\left\{{}\begin{matrix}x=2k\\y=3k\end{matrix}\right.\)
Thay vào M ta được :
\(M=\dfrac{5\cdot2k+3\cdot3k}{6\cdot2k-7\cdot3k}=\dfrac{k\left(10+9\right)}{k\left(12-21\right)}=\dfrac{-19}{9}\)