Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Coi đề lại câu a
b,
\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\\ \dfrac{x-1}{2}=\dfrac{2\left(y-2\right)}{2\cdot3}=\dfrac{3\cdot\left(z-3\right)}{3\cdot4}\\ \dfrac{x-1}{2}=\dfrac{2y-4}{6}=\dfrac{3z-9}{12}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x-1}{2}=\dfrac{2y-4}{6}=\dfrac{3z-9}{12}=\dfrac{x-1-\left(2y-4\right)+3z-9}{2-6+12}=\dfrac{x-1-2y+4+3z-9}{8}=\dfrac{\left(x-2y+3z\right)+\left(4-1-9\right)}{8}=\dfrac{14+\left(-6\right)}{8}=\dfrac{8}{8}=1\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x-1}{2}=1\Rightarrow x-1=2\Rightarrow x=3\\\dfrac{2y-4}{6}=1\Rightarrow2y-4=6\Rightarrow2y=10\Rightarrow y=5\\\dfrac{3z-9}{12}=1\Rightarrow3z-9=12\Rightarrow3z=21\Rightarrow z=7\end{matrix}\right.\)
Vậy x = 3; y = 5; z = 7
\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\)
\(\Rightarrow\dfrac{x-1}{2}=\dfrac{2y-4}{6}=\dfrac{3z-9}{12}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x-1}{2}=\dfrac{2y-4}{6}=\dfrac{3z-9}{12}\)
\(=\dfrac{x-1-2y+4+3z-9}{2-6+12}\)
\(=\dfrac{14-6}{14-6}=1\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x-1}{2}=1\\\dfrac{y-2}{3}=1\\\dfrac{z-3}{4}=1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=3\\y=5\\z=7\end{matrix}\right.\)
\(\dfrac{x}{2}=\dfrac{y}{3}\) ⇒ \(\dfrac{x}{8}=\dfrac{y}{12}\) (1)
\(\dfrac{y}{4}=\dfrac{z}{5}\) ⇒ \(\dfrac{y}{12}=\dfrac{z}{15}\) (2)
Từ (1) và (2) ⇒ \(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)
\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)\(=\dfrac{x+y-z}{8+12-15}\) \(=\dfrac{10}{5}=2\)
⇒ \(\left\{{}\begin{matrix}\dfrac{x}{8}=2\\\dfrac{y}{12}=2\\\dfrac{z}{15}=2\end{matrix}\right.\) ⇒\(\left\{{}\begin{matrix}x=16\\y=24\\z=30\end{matrix}\right.\)
Ta có \(\dfrac{x}{2}=\dfrac{y}{3}\) => \(\dfrac{1}{4}\cdot\dfrac{x}{2}=\dfrac{1}{4}\cdot\dfrac{y}{3}\Rightarrow\dfrac{x}{8}=\dfrac{y}{12}\left(1\right)\)
\(\dfrac{y}{4}=\dfrac{z}{5}\Rightarrow\dfrac{1}{3}\cdot\dfrac{y}{4}=\dfrac{1}{3}\cdot\dfrac{z}{5}\Rightarrow\dfrac{y}{12}=\dfrac{z}{15}\left(2\right)\)
Từ ( 1 ) và ( 2 ) ta có
\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\) và x+y-z=10
Áp dụng t/c dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x+y-z}{8+12-15}=\dfrac{10}{5}=2\)
\(\Rightarrow\dfrac{x}{8}=2\Rightarrow x=2\cdot8=16\)
\(\dfrac{y}{12}=2\Rightarrow=2\cdot12=24\)
\(\dfrac{z}{15}=2\Rightarrow z=2\cdot15=30\)
vậy x = 16; y = 24; z = 30
Chúc bn học tốt 
a: 2x-3y-4z=24
Áp dụng tính chất của DTSBN, ta được:
\(\dfrac{x}{1}=\dfrac{y}{6}=\dfrac{z}{3}=\dfrac{2x-3y-4z}{2\cdot1-3\cdot6-4\cdot3}=\dfrac{24}{-28}=\dfrac{-6}{7}\)
=>x=-6/7; y=-36/7; z=-18/7
b: 6x=10y=15z
=>x/10=y/6=z/4=k
=>x=10k; y=6k; z=4k
x+y-z=90
=>10k+6k-4k=90
=>12k=90
=>k=7,5
=>x=75; y=45; z=30
d: x/4=y/3
=>x/20=y/15
y/5=z/3
=>y/15=z/9
=>x/20=y/15=z/9
Áp dụng tính chất của DTSBN, ta được:
\(\dfrac{x}{20}=\dfrac{y}{15}=\dfrac{z}{9}=\dfrac{x-y-z}{20-15-9}=\dfrac{-100}{-4}=25\)
=>x=500; y=375; z=225
\(a,\dfrac{x}{5}=\dfrac{y}{6};\dfrac{y}{8}=\dfrac{x}{7}\) và \(x+y+z=138\)
\(\dfrac{x}{5}=\dfrac{y}{6}\Leftrightarrow\dfrac{x}{20}=\dfrac{y}{24}\) \(\left(1\right)\)
\(\dfrac{y}{8}=\dfrac{z}{7}\Leftrightarrow\dfrac{y}{24}=\dfrac{z}{21}\) \(\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\) \(\Leftrightarrow\dfrac{x}{20}=\dfrac{y}{24}=\dfrac{z}{21}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{x}{20}=\dfrac{y}{24}=\dfrac{z}{21}=\dfrac{x+y+z}{20+24+21}=\dfrac{138}{65}\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{20}=\dfrac{138}{65}\\\dfrac{y}{24}=\dfrac{138}{65}\\\dfrac{z}{21}=\dfrac{138}{65}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{553}{13}\\y=\dfrac{3312}{65}\\z=\dfrac{2898}{65}\end{matrix}\right.\)
Vậy.......
a)
Ta có: \(9x=5y=15z\Rightarrow\dfrac{9x}{45}=\dfrac{5y}{45}=\dfrac{15z}{45}\Rightarrow\dfrac{x}{5}=\dfrac{y}{9}=\dfrac{z}{3}\Rightarrow\dfrac{-x}{-5}=\dfrac{y}{9}=\dfrac{z}{3}_{\left(1\right)}\)
và \(-x+y-z=11_{\left(2\right)}.\)
Từ \(_{\left(1\right)}\) và \(_{\left(2\right)}\), kết hợp tính chất dãy tỉ só bằng nhau có:
\(\dfrac{-x}{-5}=\dfrac{y}{9}=\dfrac{z}{3}=\dfrac{-x+y-z}{-5+9-3}=\dfrac{11}{1}=11.\)
Từ đó: \(\left\{{}\begin{matrix}\dfrac{-x}{-5}=11\Rightarrow-x=-55\Rightarrow x=55.\\\dfrac{y}{9}=11\Rightarrow y=99.\\\dfrac{z}{3}=11\Rightarrow z=33.\end{matrix}\right.\)
Vậy.....
b); c); d); e) làm tương tự.
Ta có:
\(\dfrac{x}{2x+y+z}=\dfrac{x}{\left(x+y\right)+\left(y+z\right)}\le\dfrac{x}{2\sqrt{\left(x+y\right)\left(y+z\right)}}\)
Tương tự với các phân số khác
\(\Rightarrow VT\le\dfrac{1}{2}\left(\dfrac{x}{\sqrt{\left(x+y\right)\left(z+x\right)}}+\dfrac{y}{\sqrt{\left(y+z\right)\left(x+y\right)}}+\dfrac{z}{\sqrt{\left(z+x\right)\left(x+y\right)}}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{\sqrt{x}\cdot\sqrt{x}}{\sqrt{x+y}\cdot\sqrt{z+x}}+\dfrac{\sqrt{y}\cdot\sqrt{y}}{\sqrt{y+z}\cdot\sqrt{x+y}}+\dfrac{\sqrt{z}\cdot\sqrt{z}}{\sqrt{z+x}\cdot\sqrt{y+z}}\right)\)
\(\le\dfrac{1}{2}\left(\dfrac{\dfrac{x}{x+y}+\dfrac{x}{z+x}}{2}+\dfrac{\dfrac{y}{y+z}+\dfrac{y}{x+y}}{2}+\dfrac{\dfrac{z}{z+x}+\dfrac{z}{y+z}}{2}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{\left(\dfrac{x}{x+y}+\dfrac{y}{x+y}\right)+\left(\dfrac{y}{y+z}+\dfrac{z}{y+z}\right)+\left(\dfrac{z}{z+x}+\dfrac{x}{z+x}\right)}{2}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{3}{2}=\dfrac{3}{4}\)
Dấu "=" xảy ra khi x = y = z
\(\frac{x+1}{3}=\frac{y-2}{5}=\frac{2z+14}{9}\)
\(\Leftrightarrow\frac{2x+2}{6}=\frac{2y-4}{10}=\frac{2z+14}{9}\)
\(=\frac{2x+2-\left(2y-4\right)+2z+14}{6-10+9}=\frac{\left(2x+2z-2y\right)+20}{5}\)(Dãy tỉ số bằng nhau)
Ta có: \(x+z=y\Leftrightarrow2\left(x+z\right)=2y\)
\(\Leftrightarrow2x+2z=2y\Leftrightarrow2x+2z-2y=0\)
\(\Rightarrow\frac{\left(2x+2x-2y\right)+20}{5}=\frac{20}{5}=4\)
\(\Leftrightarrow\frac{2x+2}{6}=\frac{2y-4}{10}=\frac{2z+14}{9}=4\)
\(\Leftrightarrow\hept{\begin{cases}2x+2=24\\2y-4=40\\2z+14=36\end{cases}\Leftrightarrow\hept{\begin{cases}2x=22\\2y=44\\2z=22\end{cases}}\Leftrightarrow}\hept{\begin{cases}x=11\\y=22\\z=11\end{cases}}\)
Vậy \(x=z=11;y=22.\)
Áp dụng tính chất dãy tỉ số bằng nhau được:
\(\dfrac{x}{2x+y+z}\)=\(\dfrac{y}{2y+x+z}\)=\(\dfrac{z}{2z+x+y}\)=\(\dfrac{x+y+z}{2x+y+z+2y+x+z+2z+x+y}\)=\(\dfrac{x+y+z}{3x+3y+3z}\)=\(\dfrac{x+y+z}{3.\left(x+y+z\right)}\)=\(\dfrac{1}{3}\)=\(\dfrac{3}{9}\)<\(\dfrac{3}{4}\)(đpcm)
a, \(\frac{2}{3}x=\frac{3}{4}y=\frac{4}{5}z\)
\(\Rightarrow\frac{2x}{3.12}=\frac{3y}{4.12}=\frac{4z}{5.12}\)
\(\Rightarrow\frac{x}{18}=\frac{y}{16}=\frac{z}{15}=\frac{x+y+z}{18+16+15}=\frac{45}{49}\)
Đến đây tự làm tiếp nhé
b, \(2x=3y=5z\Rightarrow\frac{2x}{30}=\frac{3y}{30}=\frac{5z}{30}\Rightarrow\frac{x}{15}=\frac{y}{10}=\frac{z}{6}=\frac{x+y-z}{15+10-6}=\frac{95}{19}=5\)
=> x = 75, y = 50, z = 30
c, \(\frac{3}{4}x=\frac{5}{7}y=\frac{10}{11}z\)
\(\Rightarrow\frac{3x}{4.30}=\frac{5y}{7.30}=\frac{10z}{11.30}\)
\(\Rightarrow\frac{x}{40}=\frac{y}{42}=\frac{z}{33}\)
\(\Rightarrow\frac{2x}{80}=\frac{3y}{126}=\frac{4z}{132}=\frac{2x-3y+4z}{80-126+132}=\frac{8,6}{86}=\frac{1}{10}\)
=> x=... , y=... , z=...
d, Đặt \(\frac{x}{2}=\frac{y}{5}=k\Rightarrow x=2k,y=5k\)
Ta có: xy = 90 => 2k.5k = 90 => 10k2 = 90 => k2 = 9 => k = 3 hoặc -3
Với k = 3 => x = 6, y = 15
Với k = -3 => x = -6, y = -15
Vậy...
e, Tương tự câu d
b) Ta có :\(\text{ 2x = 3y = 5z }=\frac{x}{\frac{1}{2}}=\frac{y}{\frac{1}{3}}=\frac{z}{\frac{1}{5}}=\frac{x+y-z}{\frac{1}{2}+\frac{1}{3}-\frac{1}{5}}=\frac{95}{\frac{19}{30}}=\frac{1}{6}\)
=> \(2x=\frac{1}{6}\Rightarrow x=\frac{1}{12}\)
\(3y=\frac{1}{6}\Rightarrow y=\frac{1}{18}\)
\(5z=\frac{1}{6}\Rightarrow z=\frac{1}{30}\)
Theo bài ra ta có: x/3=y/4=z/6 và x-y+2z =121
\(\Rightarrow\)z/6 = 2z/12
\(\Rightarrow\)x/3=y/4=2z/12
Áp dụng tính chất dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{3}\)= \(\dfrac{y}{4}\)=\(\dfrac{2z}{12}\)= \(\dfrac{x-y+2z}{3-4+12}\)= \(\dfrac{121}{11}\)=11
+ \(\dfrac{x}{3}\)=11\(\Rightarrow\) x = 11 . 3 = 33
+ \(\dfrac{y}{4}\)=11 \(\Rightarrow\)y = 11.4 = 44
+\(\dfrac{2z}{12}\)=11 \(\Rightarrow\)2z = 11 . 12 =132
\(\Rightarrow\)z = 132 : 2 = 66
Vậy x = 33 ; y = 44 ; z = 66.
Ko có gì đâu bạn