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Ta có : \(\frac{x}{4}=\frac{y}{7}\)
\(\Rightarrow\frac{x^2}{16}=\frac{y^2}{49}=\frac{3x^2}{48}=\frac{4y^2}{196}=\frac{3x^2-4y^2}{48-196}=\frac{100}{-148}=-\frac{25}{37}\)
Thay vào là ra nhé !:D
Cái chỗ Nguyễn Quang Trung đúng ròi
\(\Rightarrow\hept{\begin{cases}\frac{x}{4}=-\frac{25}{37}\\\frac{y}{7}=-\frac{25}{37}\end{cases}}\Rightarrow\hept{\begin{cases}x=-\frac{100}{37}\\y=-\frac{175}{37}\end{cases}}\)
A) x/y-3/8=1 x/y-3/8=1/2 B)4/9:x/y=1 4/9:x/y=2/3
x/y=1+3/8 x/y=1/2+3/8 x/y=4/9:1 x/y=4/9:2/3
x/y=8/8+3/8 x/y=4/8+3/8 x/y=4/9 x/y=2/3
x/y=11/8 x/y=7/8
bài 1
a, X/Y x 1/5=3/10
X/Y=3/10:1/5
X/Y=3/2
b, 3/4 x X/Y= 6/5
X/Y=3/4:6/5
X/Y=5/8
Bài 2:
a, > b, < c,= d,<
c)\(\dfrac{3}{8}\times\dfrac{5}{8}+y=\dfrac{5}{4}\)
\(\dfrac{15}{64}+y=\dfrac{5}{4}\)
\(y=\dfrac{5}{4}-\dfrac{15}{64}\)
\(y=\dfrac{65}{64}\)
d, \(\dfrac{3}{8}+\dfrac{5}{8}\times y=\dfrac{5}{4}\)
\(\dfrac{5}{8}\times y=\dfrac{5}{4}-\dfrac{3}{8}\)
\(\dfrac{5}{8}\times y=\dfrac{7}{8}\)
\(y=\dfrac{7}{8}:\dfrac{5}{8}\)
\(y=\dfrac{7}{5}\)
a, 3/4 x y = 3/5 + 3/10
3/4 x y = 9/10
y = 9/10 : 3/4
y = 6/5
b, 3/5 : y = 3/4 - 2/5
3/5 : y = 7/20
y = 3/5 : 7/20
y = 12/7
Bài 1:
Tổng của 2 số là
\(36\times2=72\)
Số lớn là
\(72-17=55\)
Bài 2:
a) \(4567+y\div34=10987\)
\(y\div34=10987-4567\)
\(y\div34=6420\)
\(y=6420\times34\)
\(y=218280\)
b) \(\dfrac{4}{3}+\dfrac{1}{2}\div y=2\)
\(\dfrac{1}{2}\div y=2-\dfrac{4}{3}\)
\(\dfrac{1}{2}\div y=\dfrac{2}{3}\)
\(y=\dfrac{1}{2}\div\dfrac{2}{3}\)
\(y=\dfrac{3}{4}\)
Bài 3:
a) \(\dfrac{2}{5}\times\dfrac{2}{5}+\dfrac{9}{8}\div3=\dfrac{4}{25}+\dfrac{9}{8}\times\dfrac{1}{3}=\dfrac{4}{25}+\dfrac{3}{8}=\dfrac{107}{200}\)
b) \(2-\left(\dfrac{1}{7}\times4+\dfrac{5}{21}\right)=2-\left(\dfrac{4}{7}+\dfrac{5}{21}\right)=2-\dfrac{17}{21}=\dfrac{25}{21}\)
Bài 1 : Gọi a là số lớn, b là số bé, theo đề bài ta có :
(a+b):2=36⇒a+b=72
mà b=17
Nên a=72-17=55
Bài 2 :
a) 4567+y:34=10987
⇒ y:34=10987-4567
⇒ y:34=6420
⇒ y=6420x34
⇒ y=218280
b) \(\dfrac{4}{3}+\dfrac{1}{2}:y=2\)
\(\Rightarrow\dfrac{1}{2}:y=2-\dfrac{4}{3}\)
\(\Rightarrow\dfrac{1}{2}:y=\dfrac{2}{3}\)
\(\Rightarrow y=\dfrac{1}{2}:\dfrac{2}{3}\)
\(\Rightarrow y=\dfrac{1}{2}x\dfrac{3}{2}\)
\(\Rightarrow y=\dfrac{3}{4}\)
Bài 3 :
\(\dfrac{2}{5}x\dfrac{2}{5}+\dfrac{9}{8}:3=\dfrac{4}{25}+\dfrac{9}{8}x\dfrac{1}{3}=\dfrac{4}{25}+\dfrac{3}{8}\)
= \(\dfrac{4x8}{25x8}+\dfrac{25x3}{25x8}=\dfrac{32}{200}+\dfrac{75}{200}=\dfrac{107}{200}\)
\(2-\left(\dfrac{1}{7}x4+\dfrac{5}{21}\right)=2-\left(\dfrac{4}{7}+\dfrac{5}{21}\right)=2-\left(\dfrac{12}{21}+\dfrac{5}{21}\right)=2-\dfrac{17}{21}=\dfrac{42}{21}-\dfrac{17}{21}=\dfrac{25}{21}\)
1/2-2y=9/20
=>2y=1/2-9/20=1/20
=>y=1/20:2=1/40
b,3/5:4/3:y=2+7/10=9/20:y=27/10
=>y=9/20:27/10=1/6
c,y+y*3/2-y*1/2=1/10
=>y(1+3/2-1/2)=1/10
=>2y=1/10
=>y=1/10:2=1/20
\(a,\Rightarrow9y+280=1225\\ \Rightarrow9y=945\\ \Rightarrow y=105\\ b,\Rightarrow y-798=188\\ \Rightarrow y=986\\ c,\Rightarrow y:5=628\\ \Rightarrow y=3140\)
a: \(9y+35\cdot8=1225\)
\(\Leftrightarrow9y=945\)
hay y=105
b: Ta có: \(y-378-420=188\)
\(\Leftrightarrow y=188+420+378\)
hay y=986
c: Ta có: \(954-\left(y:5\right)=326\)
\(\Leftrightarrow y:5=628\)
hay y=3140