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a) \(\frac{2x+3}{5x+2}=\frac{4x+5}{10x+2}\)
\(\Leftrightarrow\)\(\left(2x+3\right)\left(10x+2\right)=\left(5x+2\right)\left(4x+5\right)\)
\(\Leftrightarrow20x^2+4x+30x+6=20x^2+25x+8x+10\)
\(\Leftrightarrow20x^2-20x^2+4x+30x-25x-8x=10-6\)
\(\Leftrightarrow x=4\)
b) \(\frac{3x-1}{40-5x}=\frac{25-3x}{5x-34}\)
\(\Leftrightarrow\left(3x-1\right)\left(5x-34\right)=\left(40-5x\right)\left(25-3x\right)\)
\(\Leftrightarrow15x^2-102x-5x+34=1000-120x-125x+15x^2\)
\(\Leftrightarrow15x^2-15x^2-102x-5x+120x+125x=1000-34\)
\(\Leftrightarrow138x=966\)
\(\Leftrightarrow x=7\)
a ) \(\frac{2x+3}{5x+2}=\frac{4x+5}{10x+2}\)
\(\left(2x+3\right).\left(10x+2\right)=\left(5x+2\right)\left(4x+5\right)\)
\(20x^2+4x+30x+6=20x^2+25x+8x+10\)
\(4x+30x-25x-8x=10-6\)
\(x=4\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\frac{2x+3}{5x+2}=\frac{4x+5}{10x+2}=\frac{2.\left(2x+3\right)-\left(4x+5\right)}{2.\left(5x+2\right)-\left(10x+2\right)}=\frac{4x+6-4x-5}{10x+4-10x-2}=\frac{1}{2}\)
Suy ra:
\(\frac{2x+3}{5x+2}=\frac{1}{2}\Rightarrow2.\left(2x+3\right)=1.\left(5x+2\right)\Rightarrow4x+6=5x+2\)
\(\Rightarrow x=4\)
a)\(\left(\frac{4}{5}\right)^{2x+7}=\left(\frac{4}{5}\right)^4\)
=> 2x + 7 = 4
2x = 4 - 7
2x = -3
x = -3 : 2
x = -1,5
Vậy x = -1,5
a ) \(\frac{2x+3}{5x+2}=\frac{4x+5}{10x+2}\)
\(\Leftrightarrow\left(2x+3\right).\left(10x+2\right)=\left(5x+2\right).\left(4x+5\right)\)
\(\Leftrightarrow20x^2+4x+30x+6=20x^2+25x+8x+10\)
\(\Leftrightarrow4x+30x-25x-8x=10-6\)
\(\Leftrightarrow1x=4\)
\(\Leftrightarrow x=4:1\)
\(\Leftrightarrow x=4\)
b ) \(\frac{3x-1}{40-5x}=\frac{25-3x}{5x-34}\)
\(\Leftrightarrow\left(3x-1\right).\left(5x-34\right)=\left(40-5x\right).\left(25-3x\right)\)
\(\Leftrightarrow15x^2-102x-5x+34=1000-120x-125x+15x^2\)
\(\Leftrightarrow-102x-5x+120x+125x=1000-34\)
\(\Leftrightarrow138x=966\)
\(\Leftrightarrow x=966:138\)
\(\Leftrightarrow x=7\)
2a) Áp dụng t/c của dãy tỉ số bằng nhau, ta có:
\(\frac{x}{10}=\frac{y}{6}=\frac{z}{21}\) => \(\frac{5x}{50}=\frac{y}{6}=\frac{2z}{42}=\frac{5x+y-2z}{50+6-42}=\frac{28}{14}=2\)
=> \(\hept{\begin{cases}\frac{x}{10}=2\\\frac{y}{6}=2\\\frac{z}{21}=2\end{cases}}\) => \(\hept{\begin{cases}x=2.10=20\\y=2.6=12\\z=2.21=42\end{cases}}\)
Vậy x,y,z lần lượt là 20; 12; 42
#)Giải :
Bài 2 :
d) Đặt \(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}=k\)
\(\Rightarrow x=2k;y=3k;z=5k\)
\(\Rightarrow2k.3k.5k=810\)
\(\Rightarrow30k^3=810\)
\(\Rightarrow k^3=3\)
\(\Rightarrow k=3\)
\(\Rightarrow\hept{\begin{cases}\frac{x}{2}=3\\\frac{y}{3}=3\\\frac{z}{5}=3\end{cases}\Rightarrow\hept{\begin{cases}x=6\\x=9\\x=15\end{cases}}}\)
Vậy x = 6; y = 9; z = 15
a/
\(\frac{x}{10}=\frac{y}{6}=\frac{z}{21}=\frac{5x}{50}=\frac{y}{6}=\frac{2z}{42}\)\(=\frac{5x+y-2z}{50+6-42}=\frac{28}{14}=2\)\(\Rightarrow x=20;y=12;z=42\)
b/\(3x=2y\Leftrightarrow\frac{x}{2}=\frac{y}{3};7y=5z\Leftrightarrow\frac{y}{5}=\frac{z}{7}\)\(\Rightarrow\frac{x}{10}=\frac{y}{15}=\frac{z}{21}=\frac{x-y+z}{10-15+20}=2\)
\(\Rightarrow x=20;y=30;z=42\)
\(a.\frac{x}{3}=\frac{y}{4};\frac{y}{5}=\frac{z}{7}\) và \(2x+3y-z=186\)
Từ \(\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{x}{3}\times\frac{1}{5}=\frac{y}{4}\times\frac{1}{5}=\frac{x}{15}=\frac{y}{20}\left(1\right)\)
Từ \(\frac{y}{5}=\frac{z}{7}\Rightarrow\frac{y}{5}\times\frac{1}{4}=\frac{z}{7}\times\frac{1}{4}=\frac{y}{20}=\frac{z}{28}\left(2\right)\)
Từ \(\left(1\right)\)và \(\left(2\right)\)\(\Rightarrow\)\(\frac{x}{15}=\frac{y}{20}=\frac{z}{28}\)
Đặt \(\frac{x}{15}=\frac{y}{20}=\frac{z}{28}=k\)
\(\Rightarrow\hept{\begin{cases}x=15k\\y=20k\\z=28k\end{cases}}\)
Lại có : \(2x+3y-z=186\)
Thay vào ta có :
\(2.15k+3.20k-28k=186\)
\(30k+60k-28k=186\)
\(62k=186\)
\(k=3\)
Thay vào ta được :
\(\Rightarrow\hept{\begin{cases}x=15.3=45\\y=20.3=60\\z=28.3=84\end{cases}}\)
Vậy .....
ta có: 2xx=3y=>x/3=y/2=>x/21=y/14 ; x/7=z/5=>x/21=z/15 =>x/21=y/14=z/15=>3x/63=7y/98=5z/75 ADTCDTSBN ta có 3x/63=7y/98=5z /75=3x-7y+5z=40/63-98+75=40=1 3x=1.63=63 =>x=21 ;7y=1.98=98=>y=14 ; 5z=1.75=>z=15
