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1) Áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\frac{12x-15y}{7}=\frac{20y-12x}{9}=\frac{15y-20z}{11}=\frac{12x-15y+20z-12x+15y-20z}{7+9+11}=\frac{0}{27}=0\)
\(\Rightarrow\hept{\begin{cases}12x-15y=0\\15y-20z=0\end{cases}\Rightarrow}\hept{\begin{cases}12x=15y\\15y=20z\end{cases}\Rightarrow\hept{\begin{cases}\frac{x}{15}=\frac{y}{12}\\\frac{y}{20}=\frac{z}{15}\end{cases}\Rightarrow}\hept{\begin{cases}\frac{x}{75}=\frac{y}{60}\\\frac{y}{60}=\frac{z}{45}\end{cases}\Rightarrow}\frac{x}{75}=\frac{y}{60}=\frac{z}{45}}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\frac{x}{75}=\frac{y}{60}=\frac{z}{45}=\frac{x+y+z}{75+60+45}=\frac{48}{180}=\frac{4}{15}\)
=> x = 75.4 : 15 = 20 ;
y = 60.4 : 15 = 16 ;
z = 45.4 : 15 = 12
Vậy x = 20 ; y = 16 ; z = 12
2) Từ đẳng thức \(\frac{x}{y+z+t}=\frac{y}{z+t+x}=\frac{z}{t+x+y}=\frac{t}{x+y+z}\)
\(\Rightarrow\frac{z}{y+z+t}+1=\frac{y}{z+t+x}+1=\frac{z}{t+x+y}+1=\frac{t}{x+y+z}+1\)
\(\Rightarrow\frac{x+y+z+t}{y+z+t}=\frac{x+y+z+t}{z+t+x}=\frac{x+y+z+t}{t+x+y}=\frac{x+y+z+t}{x+y+z}\)
Nếu x + y + z + t = 0
=> x + y = - (z + t)
=> y + z = - (t + x)
=> z + t = - (x + y)
=> t + x = - (z + y)
Khi đó :
P = \(\frac{-\left(z+t\right)}{z+t}+\frac{-\left(t+x\right)}{t+x}+\frac{-\left(x+y\right)}{x+y}+\frac{-\left(z+y\right)}{z+y}=-1+\left(-1\right)+\left(-1\right)+\left(-1\right)=-4\)
=> P = 4
Nếu x + y + z + t khác 0
=> \(\frac{1}{y+z+t}=\frac{1}{z+t+x}=\frac{1}{t+x+y}=\frac{1}{x+y+z}\)
=> y + z + t = z + t + x = t + x + y = x + y + z
=> x =y = z = t
Khi đó : P = 1 + 1 + 1 + 1 = 4
Vậy nếu x + y + z + t = 0 thì P = - 4
nếu x + y + z + t khác 0 thì P = 4
\(\frac{2^{12}.3^5-4^6.81}{\left(2^2.3\right)^6+8^4.3^5}\)
\(=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}\)
\(=\frac{2^{12}.\left(3^5-3^4\right)}{2^{12}.\left(3^6+3^5\right)}\)
\(=\frac{3^5-3^4}{3^6+3^5}=\frac{3^4.\left(3-1\right)}{3^5\left(3+1\right)}\)
\(=\frac{3^4.2}{3^5.4}=\frac{3^4.2}{3^4.3.4}=\frac{2}{12}=\frac{1}{6}\)
P/s: Hoq chắc ạ (: Ms lp 6 lm đại
\(\frac{x}{2}=\frac{y}{3}\)
\(\Leftrightarrow\frac{x}{8}=\frac{y}{12}\)(1)
\(\frac{y}{4}=\frac{z}{5}\)
\(\Leftrightarrow\frac{y}{12}=\frac{z}{15}\)(2)
Từ (1) (2)
\(\Rightarrow\frac{x}{8}=\frac{y}{12}=\frac{z}{15}=\frac{x+y-z}{8+12-15}=\frac{10}{5}=2\)
\(\Rightarrow\hept{\begin{cases}x=2.8\\y=2.12\\z=2.15\end{cases}\Rightarrow}\hept{\begin{cases}x=16\\y=24\\z=30\end{cases}}\)
A=\(\left(\frac{1}{4}-1\right).\left(\frac{1}{9}-1\right).\left(\frac{1}{16}-1\right).............\left(\frac{1}{9801}-1\right).\left(\frac{1}{10000}-1\right)\)
A=\(\left(\frac{1-4}{4}\right).\left(\frac{1-9}{9}\right).\left(\frac{1-16}{16}\right).............\left(\frac{1-9801}{9801}\right).\left(\frac{1-10000}{10000}\right)\)
A=\(\frac{-3}{4}.\frac{-8}{9}.\frac{-15}{16}.....................\frac{-9800}{9801}.\frac{-9999}{10000}\)
A=\(\frac{-1.3}{2^2}.\frac{-2.4}{3^2}.\frac{-3.5}{4^2}.....................\frac{-98.100}{99^2}.\frac{-99.101}{100^2}\)
A=\(\frac{\left[\left(-1\right).\left(-2\right).\left(-3\right)....................\left(-98\right).\left(-99\right)\right].\left(3.4.5............100.101\right)}{\left(2.3.4.........99.100\right).\left(2.3.4...............99.100\right)}\)
A=\(\frac{1.101}{100.2}\)=\(\frac{101}{200}\)
2
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.................+\frac{2}{x.\left(x+1\right)}=\frac{2015}{2017}\)
\(\frac{1}{3.2}+\frac{1}{6.2}+\frac{1}{10.2}+.................+\frac{2}{2.x.\left(x+1\right)}=\frac{1}{2}.\frac{2015}{2017}\)
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+.................+\frac{1}{x.\left(x+1\right)}=\frac{2015}{2017}.\frac{1}{2}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+..................+\frac{1}{x.\left(x+1\right)}=\frac{2015}{2017}.\frac{1}{2}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+..............+\frac{1}{x}-\frac{1}{x+1}=\frac{2015}{2017}.\frac{1}{2}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2015}{2017}.\frac{1}{2}\)
\(\frac{x+1}{2.\left(x+1\right)}-\frac{2}{2.\left(x+1\right)}=\frac{2015}{2017}.\frac{1}{2}\)
\(\frac{\left(x+1\right)-2}{2.\left(x+1\right)}=\frac{2015}{2017}.\frac{1}{2}\)
\(\frac{x-1}{2.\left(x+1\right)}=\frac{2015}{2017}.\frac{1}{2}\)
=>\(\frac{x-1}{x+1}=\frac{2015}{2017}.\frac{1}{2}:\frac{1}{2}\)
\(\frac{x-1}{x+1}=\frac{2015}{2017}\)
=>x+1=2017
=>x=2018-1
=>x=2016
Vậy x=2016
Còn bài 3 em ko biết làm em ms lớp 6
Chúc anh học tốt
Câu 1 : Ta có :
\(\hept{\begin{cases}\left|x+y-5\right|\ge0\forall x;y\\\left|2x-y+8\right|\ge0\forall x;y\end{cases}\Rightarrow\left|x+y-5\right|+\left|2x-y+8\right|\ge0\forall x;y}\)
Dấu \("="\)xảy ra
\(\Leftrightarrow\hept{\begin{cases}\left|x+y-5\right|=0\\\left|2x-y+8\right|=0\end{cases}\Leftrightarrow\hept{\begin{cases}x+y-5=0\\2x-y+8=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x+y=5\\2x-y=-8\end{cases}}}\)
\(\Leftrightarrow x+y+2x-y=5+-8\)
\(\Leftrightarrow3x=-3\)
\(\Leftrightarrow x=-1\)
Mà \(x+y=5\Rightarrow y=5-\left(-1\right)=6\)
Vậy \(x=-1;y=6\)
Câu 2 : Ta có :
\(\left|x\right|\ge0\forall x;\left|x+2\right|\ge0\forall x\)
\(\Rightarrow\left|x\right|+\left|x+2\right|\ge0\forall x\)
Dấu \("="\)xảy ra
\(\Leftrightarrow\hept{\begin{cases}\left|x\right|=0\\\left|x+2\right|=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\x=-2\end{cases}\Leftrightarrow}}\)Loại
Vậy không có TH x thỏa mãn
Câu 3 : Ta có :
\(\left|-y\right|\ge0\forall y\)
\(\Rightarrow\frac{-2}{5}-\left|-y\right|\le-\frac{2}{5}\)
Mà : \(\left|\frac{1}{2}-\frac{1}{3}+x\right|\ge0\forall x\)
\(\Rightarrow\left|\frac{1}{2}-\frac{1}{3}+x\right|=-\frac{2}{5}-\left|-y\right|\)( vô lý )
Vậy không có TH x thỏa mãn
Bài 2 :
\(\frac{-7}{6}=\frac{x}{8}\)\(\Rightarrow x=\frac{-7.8}{6}=\frac{-28}{3}\)
\(\frac{-7}{6}=\frac{-98}{y}\)\(\Rightarrow y=\frac{6.\left(-98\right)}{-7}=84\)
\(\frac{-7}{6}=\frac{-14}{z}\)\(\Rightarrow z=\frac{6.\left(-14\right)}{-7}=12\)
\(\frac{-7}{6}=\frac{t}{102}\)\(\Rightarrow t=\frac{\left(-7\right).102}{6}=-119\)
\(\frac{-7}{6}=\frac{u}{-78}\)\(\Rightarrow u=\frac{\left(-7\right).\left(-78\right)}{6}=91\)