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Từ \(\dfrac{x+4}{7+y}=\dfrac{4}{7}\Rightarrow\dfrac{x+4}{4}=\dfrac{y+7}{7}\)
Áp dụng t/c của dãy tỉ số bằng nhau, ta có:
\(\dfrac{x+4}{4}=\dfrac{y+7}{7}=\dfrac{x+y+11}{11}=\dfrac{22+11}{11}=\dfrac{33}{11}=3\\ \Rightarrow x=3.4-4=8\\ y=3.7-7=14\)
Từ x + y = 22 => y=22-x
Thay y=22-x vào biểu thức ta có:
\(\dfrac{x+4}{7+22-x}=\dfrac{4}{7}\)
\(\dfrac{x+4}{29-x}=\dfrac{4}{7}\)
\(7\left(x+4\right)=4\left(29-x\right)\)
\(7x+28=116-4x\)
\(7x+4x=116-28\)
\(11x=88\)
\(x=8\)
\(\Rightarrow y=14\)
vậy x=8 ; y=14
Lời giải:
Ta có: \(\frac{x+4}{7+y}=\frac{4}{7}\Leftrightarrow 7(x+4)=4(7+y)\)
\(\Leftrightarrow 7x=4y\)
\(\Leftrightarrow 11x=4y+4x=4(x+y)=4.22=88\)
\(\Leftrightarrow x=\frac{88}{11}=8\)
Suy ra \(y=22-x=22-8=14\)
Vậy \((x,y)=(8,14)\)
Từ \(\dfrac{x+4}{7+y}=\dfrac{4}{7}\) => ( x + 4) . 7 = (7 + y) . 4
=> 7x + 28 = 28 + 4y
=> 7x = 4y
=> \(\dfrac{x}{4}=\dfrac{y}{7}\) = \(\dfrac{x+y}{4+7}\) = \(\dfrac{22}{11}=2\) ( Áp dụng t/c dãy tỉ số bằng nhau)
=> \(\left\{{}\begin{matrix}\dfrac{x}{4}=2\\\dfrac{y}{7}=2\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x=8\\y=14\end{matrix}\right.\)
Vậy x = 8 , y = 14
a)ta có 4+x/7+y=4/7
<=>7x+28=28+4y
<=> 7x=4y
lại có x+y=22
=>4/7y+y=22
<=>11/7y=22 <=> y=14
<=> x= 4/7*14=8
vậy x=8, y=14
b) Từ x/3=y/4 va y/5=z/6-->x/15=y/20=z/24 (1)
(1) = 2x/30=3y/60=4z/96=(2x+3y+4z)/186 (2) (t/c dãy tỉ số bằng nhau)
Ta lại có
(1) = 3x/45=4y/80=5z/120=(3x+4y+5z)/245 (3)(t/c dãy tỉ số bằng nhau)
Từ (2)(3) ta có(2x+3y+4z)/186=(3x+4y+5z)/245
Vậy M = (2x+3y+4z)/(3x+4y+5z)=186/245
a/ Do \(x+y=22\Rightarrow y=22-x\)
\(\Rightarrow\dfrac{4+x}{7+22-x}=\dfrac{4}{7}\Leftrightarrow\dfrac{4+x}{29-x}=\dfrac{4}{7}\)
\(\Leftrightarrow7\left(4+x\right)=4\left(29-x\right)\Leftrightarrow28+7x=116-4x\)
\(\Leftrightarrow11x=88\Rightarrow x=8\)
\(\Rightarrow y=22-x=14\)
b/ \(\dfrac{x}{3}=\dfrac{y}{4}\Rightarrow y=\dfrac{4x}{3}\)
\(\dfrac{y}{5}=\dfrac{z}{6}\Rightarrow z=\dfrac{6y}{5}\) \(\Rightarrow z=\dfrac{6}{5}\left(\dfrac{4x}{3}\right)=\dfrac{8x}{5}\)
Vậy \(M=\dfrac{2x+3y+4z}{3x+4y+5z}=\dfrac{2x+3.\dfrac{4x}{3}+4.\dfrac{8x}{5}}{3x+4.\dfrac{4x}{3}+5.\dfrac{8x}{5}}\)
\(\Rightarrow M=\dfrac{x\left(2+4+\dfrac{32}{5}\right)}{x\left(3+\dfrac{16}{3}+8\right)}=\dfrac{\dfrac{62}{5}}{\dfrac{49}{3}}=\dfrac{186}{245}\)
Câu a:
Ta có: \(x+y=22\Rightarrow y=22-x\)
\(\Rightarrow\dfrac{4+x}{7+22-x}=\dfrac{4}{7}\Leftrightarrow\dfrac{4+x}{29-x}=\dfrac{4}{7}\)
\(\Leftrightarrow7\left(4+x\right)=4\left(29-x\right)\Leftrightarrow28+7x=116-4x\)
\(\Leftrightarrow11x=88\Rightarrow x=8\)
\(\Rightarrow y=22-x=22-8=14\)
Vậy \(x=8,y=14\)
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 :\(\text{ 2x = 3y = 5z }=\frac{x}{\frac{1}{2}}=\frac{y}{\frac{1}{3}}=\frac{z}{\frac{1}{5}}=\frac{x+y-z}{\frac{1}{2}+\frac{1}{3}-\frac{1}{5}}=\frac{95}{\frac{19}{30}}=\frac{1}{6}\)
=> \(2x=\frac{1}{6}\Rightarrow x=\frac{1}{12}\)
\(3y=\frac{1}{6}\Rightarrow y=\frac{1}{18}\)
\(5z=\frac{1}{6}\Rightarrow z=\frac{1}{30}\)
a)\(\dfrac{x}{3}=\dfrac{y}{7}=\dfrac{x+y}{3+7}=\dfrac{20}{10}=2\)
\(\dfrac{x}{3}=2\Rightarrow x=6\)
\(\dfrac{y}{7}=2\Rightarrow y=14\)
b)\(\dfrac{x}{5}=\dfrac{y}{2}=\dfrac{x-y}{5-2}=\dfrac{6}{3}=2\)
\(\dfrac{x}{5}=2\Rightarrow x=10\)
\(\dfrac{y}{2}=2\Rightarrow y=4\)
\(x+y=55\Rightarrow x=55-y\\ \Leftrightarrow\dfrac{4+55-y}{7+y}=\dfrac{4}{7}\\ \Leftrightarrow28+385-7y=28+4y\\ \Rightarrow y=35\\ \Rightarrow x=55-35=20\)
giúp
\(\dfrac{x+4}{4}=\dfrac{y+7}{7}\)Theo tc dãy tỉ số bằng nhau
\(\dfrac{x+4}{4}=\dfrac{y+7}{7}=\dfrac{x+y+4+7}{11}=\dfrac{33}{11}=3\Leftrightarrow x=8;y=14\)