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\(\left|x-\frac{1}{2}\right|+\frac{3}{4}=\left|-1,6+\frac{3}{5}\right|\)
\(\Rightarrow\left|x-\frac{1}{2}\right|+\frac{3}{4}=\left|-1,6+0,6\right|\)
\(\Rightarrow\left|x-\frac{1}{2}\right|+\frac{3}{4}=\left|-1\right|\)
\(\Rightarrow\left|x-\frac{1}{2}\right|+\frac{3}{4}=1\)
\(\Rightarrow\left|x-\frac{1}{2}\right|=1-\frac{3}{4}\)
\(\Rightarrow\left|x-\frac{1}{2}\right|=\frac{1}{4}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{2}=\frac{1}{4}\\x-\frac{1}{2}=-\frac{1}{4}\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{3}{4}\\x=\frac{1}{4}\end{cases}}}\)
Vậy ...
\(1)\) Ta có :
\(3x=4y\)\(\Leftrightarrow\)\(\frac{x}{4}=\frac{y}{3}\)\(\Leftrightarrow\)\(\frac{x}{8}=\frac{y}{6}\)
\(5y=6z\)\(\Leftrightarrow\)\(\frac{y}{6}=\frac{z}{5}\)
\(\Rightarrow\)\(\frac{x}{8}=\frac{y}{6}=\frac{z}{5}\)
Đặt \(\frac{x}{8}=\frac{y}{6}=\frac{z}{5}=k\)\(\Rightarrow\)\(\hept{\begin{cases}x=8k\\y=6k\\z=5k\end{cases}}\) \(\left(1\right)\)
Thay \(\left(1\right)\) vào \(xyz=30\) ta được :
\(8k.6k.5k=30\)
\(\Leftrightarrow\)\(240k^3=30\)
\(\Leftrightarrow\)\(k^3=\frac{30}{240}\)
\(\Leftrightarrow\)\(k^3=\frac{1}{8}\)
\(\Leftrightarrow\)\(k^3=\left(\frac{1}{2}\right)^3\)
\(\Leftrightarrow\)\(k=\frac{1}{2}\)
Suy ra :
\(x=8k=8.\frac{1}{2}=\frac{8}{2}=4\)
\(y=6k=6.\frac{1}{2}=\frac{6}{2}=3\)
\(z=5k=5.\frac{1}{2}=\frac{5}{2}\)
Vậy \(x=4\)\(;\)\(y=3\) và \(z=\frac{5}{2}\)
Chúc bạn học tốt ~
Ta có : \(3x=4y\Rightarrow\dfrac{x}{4}=\dfrac{y}{3}\Rightarrow\dfrac{x}{24}=\dfrac{y}{18}\left(1\right)\)
\(5y=6z\Rightarrow\dfrac{y}{6}=\dfrac{z}{5}\Rightarrow\dfrac{y}{18}=\dfrac{z}{15}\left(2\right)\)
Từ (1);(2) \(\Rightarrow\dfrac{x}{24}=\dfrac{y}{18}=\dfrac{z}{15}\)
Đặt \(\dfrac{x}{24}=\dfrac{y}{18}=\dfrac{z}{15}=k\Rightarrow x=24k;y=18k;z=15k\)
\(\text{Ta có }:x.y.z=24k.18k.15k=30\\ \Rightarrow k^3.6480=30\\ \Rightarrow k^3=\dfrac{1}{216}\\ \Rightarrow k=\dfrac{1}{6}\\ \Rightarrow x=24.\dfrac{1}{6}=4\\ y=18.\dfrac{1}{6}=3\\ z=15.\dfrac{1}{6}=2.5\)
Vậy x = 4 ; y = 3 ; z = 2,5
a) Tìm các số x, y, z biết rằng
3x=4y ; 5y=6z và xyz = 30
Giải
Ta có: 3x = 4y; 5y = 6z \(\Rightarrow\dfrac{x}{4}=\dfrac{y}{3}\); \(\dfrac{y}{6}=\dfrac{z}{5}\)
\(\Rightarrow\dfrac{x}{8}=\dfrac{y}{6}=\dfrac{z}{5}\)
Đặt \(\dfrac{x}{8}=\dfrac{y}{6}=\dfrac{z}{5}=k\) \(\Rightarrow\left\{{}\begin{matrix}x=8k\\y=6k\\z=5k\end{matrix}\right.\)
Có \(xyz=30\) \(\Leftrightarrow\) \(8k.6k.5k=30\)
\(\Rightarrow\) \(240k^3=30\)
\(\Rightarrow\) \(k^3=8\)
\(\Rightarrow\) \(k=\sqrt[3]{8}=2\)
\(\Rightarrow\left\{{}\begin{matrix}x=8.2=16\\y=6.2=12\\z=5.2=10\end{matrix}\right.\)
3x=4y
=>x/4=y/3
=>x/8=y/6
5y=6z
=>y/6=z/5
=>x/8=y/6=z/5
Đặt x/8=y/6=z/5=k
=>x=8k; y=6k; z=5k
xyz=30
=>8k*6k*5k=30
=>240k^3=30
=>k^3=1/8
=>k=1/2
=>x=8*1/2=4; y=6*1/2=3; z=5*1/2=5/2
a)Ta có: \(\frac{x}{y+z+1}=\frac{y}{x+y+2}=\frac{z}{x+y-3}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\frac{x}{y+z+1}=\frac{y}{x+y+2}=\frac{z}{x+y-3}\)
\(=\frac{x+y+z}{y+z+1+x+y+2+x+y-3}\)
\(=\frac{x+y+z}{2x+2y+2z}\)
\(=\frac{x+y+z}{2\left(x+y+z\right)}=\frac{1}{2}\)
b, Ta có : \(\dfrac{x}{3}=\dfrac{y}{4};\dfrac{y}{5}=\dfrac{z}{6}\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{24}\)
Đặt \(x=15k;y=20k;z=24k\)
Thay vào A ta được : \(A=\dfrac{30k+60k+96k}{45k+80k+120k}=\dfrac{186k}{245k}=\dfrac{186}{245}\)
a) \(\left|3x-\dfrac{1}{2}\right|+\left|\dfrac{1}{4}y+\dfrac{3}{5}\right|=0\)
Do \(\left|3x-\dfrac{1}{2}\right|,\left|\dfrac{1}{4}y+\dfrac{3}{5}\right|\ge0\forall x,y\)
\(\Rightarrow\left\{{}\begin{matrix}3x-\dfrac{1}{2}=0\\\dfrac{1}{4}y+\dfrac{3}{5}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{6}\\y=-\dfrac{12}{5}\end{matrix}\right.\)
b) \(\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|+\left|\dfrac{5}{7}y-\dfrac{1}{2}\right|\le0\)
Do \(\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|,\left|\dfrac{5}{7}y-\dfrac{1}{2}\right|\ge0\forall x,y\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3}{2}x+\dfrac{1}{9}=0\\\dfrac{5}{7}y-\dfrac{1}{2}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{2}{27}\\y=\dfrac{7}{10}\end{matrix}\right.\)
ta có :3x=4y,5y=6z
=>\(\dfrac{x}{4}\)=\(\dfrac{y}{3}\); \(\dfrac{y}{6}\)=\(\dfrac{z}{5}\)
=> \(\dfrac{x}{8}\)=\(\dfrac{y}{6}\); \(\dfrac{y}{6}\)=\(\dfrac{z}{5}\)
=> \(\dfrac{x}{8}\)=\(\dfrac{y}{6}\)=\(\dfrac{z}{5}\)
Đặt \(\dfrac{x}{8}\)=\(\dfrac{y}{6}\)=\(\dfrac{z}{5}\)=k
=> x=8k ; y=6k ; z=5k
=> 8k.6k.5k=30
=> 240k3 =30
=>k3 =8
=>k=2
=> x=8.2=16 ; y=6.2=12 ; x =5.2=10
ban giải cho mik các bài trc với