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Áp dụng tính chất dãy tỉ số bằng nhau ta được:
\(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}=\frac{x+y-z}{2+3-5}=\frac{10}{0}\)
Vì phân số này không có nghĩa nên bạn xem lại đề nhé
ADTCCDTS bằng nhau ta có :
\(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}\) (và x+ y -z = 10 )
\(=>\frac{x+y-z}{2+3-5}=10\)
\(=>x=10x2=20\)
\(=>y=10x3=30\)
\(=>z=10x5=50\)
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\(\text{Theo bài ra, ta suy ra:}\)
\(\text{3x= 2y; 5y= 4z.}\)
\(\text{Suy ra:}\)\(\frac{x}{y}=\frac{2}{3};\frac{y}{z}=\frac{4}{5}\)
\(\text{Suy ra:}\) \(\frac{x}{y}=\frac{8}{12};\frac{y}{z}=\frac{12}{15}\)
\(\text{Suy ra: x= 8 phần; y= 12 phần; z= 15 phần}\)
\(\text{Suy ra: x+ y- z tương ứng với: 8+12-5=5 phần. }\)
\(\text{Suy ra 1 phần tương úng với:}\)\(\text{10:5=2}\)
\(\text{Suy ra: x= 2.8=16}\)
\(\text{y=2.12=24}\)
\(\text{z=2.15=30}\)
\(\text{Vậy: x=16; y=24;z=30.}\)
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Ta có: \(\frac{6}{11}x=\frac{9}{2}y=\frac{18}{5}\)
=>\(\frac{6x}{11.18}=\frac{9y}{2.18}=\frac{18z}{5.18}=\frac{x}{33}=\frac{y}{4}=\frac{z}{5}=\frac{-x+y+z}{-33+4+5}=\frac{-120}{24}=5\)
=> x=5.33=165
y=5.4=20
z=5.5=25
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a) \(\frac{1}{2}-|\frac{5}{4}-2x|=\frac{1}{3}\Leftrightarrow|\frac{5}{4}-2x|=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{5}{4}-2x=\frac{1}{6}\\\frac{5}{4}-2x=-\frac{1}{6}\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=\frac{5}{4}-\frac{1}{6}=\frac{13}{12}\\2x=\frac{5}{4}+\frac{1}{6}=\frac{17}{12}\end{cases}}}\)
Tự làm nốt và kết luận
b) \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}+\frac{1}{14}\right)=0\)
Vì \(\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}+\frac{1}{14}\right)\ne0\forall x\Rightarrow x+1=0\Leftrightarrow x=-1\)
Vậy ....
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1,\(\frac{xyz+x+z}{yz+1}=\frac{10}{7}\Rightarrow\frac{x\left(yz+1\right)+z}{yz+1}=\frac{10}{7}\)
\(\Leftrightarrow x+\frac{z}{yz+1}=\frac{10}{7}\Leftrightarrow x+\frac{1}{\frac{yz+1}{z}}=\frac{10}{7}\)
\(\Leftrightarrow x+\frac{1}{y+\frac{1}{z}}=1+\frac{3}{7}=1+\frac{1}{\frac{7}{3}}=1+\frac{1}{2+\frac{1}{3}}\)
Nên x=1,y=2,z=3 bài này thiếu điều kiện x,y,z nhé
2,bài 2 để mai anh xem nha
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Ta có: \(\frac{2x+3}{5x+2}=\frac{4x+5}{10x+2}\)
\(\Rightarrow\left(2x+3\right).\left(10x+2\right)=\left(5x+2\right).\left(4x+5\right)\)
\(\Rightarrow20x^2+4x+30x+6=10x^2+25x+8x+10\)
\(\Rightarrow34x+6=33x+10\)
\(\Rightarrow34x-33x=-6+10\)
\(\Rightarrow x=4\)
Ta có:
\(\frac{2x+3}{5x+2}=\frac{4x+5}{10x+2}\)
\(\Rightarrow\left(2x+3\right)\left(10x+2\right)=\left(5x+2\right)\left(4x+5\right)\)
\(\Rightarrow20x^2+34x+6=20x^2+33x+10\)
\(\Rightarrow\left(20x^2+34x+6\right)-\left(20x^2+33x+6\right)=\left(20x^2+33x+10\right)-\left(20x^2+33x+6\right)\)
\(\Rightarrow\left(20x^2-20x^2\right)+\left(34x-33x\right)+\left(6-6\right)=\left(20x^2-20x^2\right)+\left(33x-33x\right)+\left(10-6\right)\)
\(\Rightarrow x=4\)
Vậy x = 4.
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Ta có \(\frac{xy}{x+y}=\frac{yz}{y+z}=\frac{xz}{x+z}\)
=> \(\frac{xyz}{xz+yz}=\frac{xyz}{xy+xz}=\frac{xyz}{xy+yz}\)
=> \(xz+yz=xy+xz=xy+yz\)(vì x ; y ;z \(\ne0\Leftrightarrow xyz\ne0\))
=> \(\hept{\begin{cases}xz+yz=xy+xz\\xy+xz=xy+yz\\xz+yz=xy+yz\end{cases}}\Rightarrow\hept{\begin{cases}yz=xy\\xz=yz\\xz=xy\end{cases}}\Rightarrow\hept{\begin{cases}z=x\\x=y\\y=z\end{cases}}\Rightarrow x=y=z\)
Khi đó M = \(\frac{x^2+y^2+z^2}{xy+yz+zx}=\frac{x^2+y^2+z^2}{x^2+y^2+z^2}=1\left(\text{vì }x=y=z\right)\)
Ơ !!??!?!!?!! chả nhìn thấy x ở đâu cả
\(\frac{10}{3}:\frac{5}{2}=\frac{10}{3}.\frac{2}{5}=\frac{20}{15}=\frac{3}{4}\)