\(\frac{72-x}{7}=\frac{x-40}{9}\)
\(\frac{x-40}{9}=\frac{72-x}{7}\)
\(\frac{x-40}{9}=\frac{72-x}{7}\)
=> 7 ( x - 40 ) = 9 ( 72 - x )
7 x - 280 = 648 - 9 x
7 x + 9 x = 648 + 280
16 x = 928
x = 58
\(\frac{x-40}{9}=\frac{72-x}{7}\)
\(\Rightarrow\left(x-40\right).7=9.\left(72-x\right)\)
\(\Rightarrow648-9x=7x-280\)
\(\Rightarrow\left(-280\right)-648=\left(-9x\right)-7x\)
\(\Rightarrow-928=-16x\)
\(\Rightarrow x=58\)
Tính giá trị biểu thức:
d) D=\(\frac{3x-2y}{x-3y}\)với \(\frac{x}{y}=\frac{10}{3}\)
Tìm x biết
\(\frac{72-x}{7}=\frac{x-40}{9}\)
(làm 2 cách
\(\text{Cách 1:}\)
\(\frac{72-x}{7}=\frac{x-40}{9}\)
=> \(\left(72-x\right)\cdot9=\left(x-40\right)\cdot7\)
=> \(648-9x=7x-280\)
=> \(648+280=7x+9x\)
=> \(16x=928\)=> \(x=928:16=58\)
\(\text{Cách 2:}\)
\(\text{Áp dụng tính chất dãy tỉ số bằng nhau, ta có:}\)
\(\frac{72-x}{7}=\frac{x-40}{9}=\frac{72-x+x-40}{7+9}=\frac{32}{16}=2\)
\(\text{Ta suy ra:}\)
\(\frac{72-x}{7}=2\)=> \(72-x=2\cdot7=14\)=> \(x=72-14=58\)
\(\text{Vậy }x=58\)
Tìm x biết:
a) \(\frac{x^2}{6}=\frac{24}{25}\)
b) \(\frac{72-x}{7}=\frac{x-40}{9}\)
a) \(\frac{x}{6}^2=\frac{24}{25}\)
\(\Rightarrow x^2.25=6.24\)
\(\Rightarrow x^2.25=144\)
\(\Rightarrow x^2=144\div25\)
\(\Rightarrow x^2=5,76=2,4^2=\left(-2,4^2\right)\)
\(\Rightarrow x\in\left\{2,4;-2,4\right\}\)
Vậy \(x\in\left\{2,4;-2,4\right\}\)
b) \(\frac{72-x}{7}=\frac{x-40}{9}\)
\(\Rightarrow\left(72-x\right).9=\left(x-40\right).7\)
\(\Rightarrow648-9x=7x-280\)
\(\Rightarrow\left(-280\right)-648\) \(=-9x-7x\)
\(\Rightarrow-928=-16x\)
\(\Rightarrow x=58\)
Vậy \(x=58\)
tính x trong các tỉ lệ thức sau : \(\frac{72-x}{7}=\frac{x-40}{9}\)
=> (72 - x) . 9 = (x - 40) . 7
=> 648 - 9x = 7x - 280
=> (-280) - 648 = -9x - 7x
-928 = -16x
=> x = 58
2, Tìm x biết :
a, \(\frac{x}{4}\) = \(\frac{3}{2}\)
b, \(\frac{x}{16}=\frac{9}{x}\)
c, \(\frac{x^2}{6}=\frac{24}{25}\)
d, \(\frac{72-9}{7}=\frac{x-40}{9}\)
a.
\(\frac{x}{4}=\frac{3}{2}\)
\(x=\frac{3}{2}\times4\)
\(x=6\)
b.
\(\frac{x}{16}=\frac{9}{x}\)
\(x\times x=16\times9\)
\(x^2=144\)
\(x^2=\left(\pm12\right)^2\)
\(x=\pm12\)
Vậy \(x=12\) hoặc \(x=-12\)
c.
\(\frac{x^2}{6}=\frac{24}{25}\)
\(x^2=\frac{24}{25}\times6\)
\(x^2=\frac{144}{25}\)
\(x^2=\left(\pm\frac{12}{5}\right)^2\)
\(x=\pm\frac{12}{5}\)
Vậy \(x=\frac{12}{5}\) hoặc \(x=-\frac{12}{5}\)
d.
\(\frac{72-9}{7}=\frac{x-40}{9}\)
\(\frac{x-40}{9}=\frac{63}{7}\)
\(x-40=\frac{63}{7}\times9\)
\(x-40=81\)
\(x=81+40\)
\(x=121\)
1 tìm x biết:
a, (2x-1) : \(1\frac{3}{7}\) = \(1\frac{13}{15}\) :\(1\frac{1}{3}\)
b, \(\frac{72-x}{7}\) = \(\frac{x-40}{9}\)
1.
b) \(\frac{72-x}{7}=\frac{x-40}{9}\)
\(\Rightarrow\left(72-x\right).9=\left(x-40\right).7\)
\(\Rightarrow648-9x=7x-280\)
\(\Rightarrow648+280=7x+9x\)
\(\Rightarrow928=16x\)
\(\Rightarrow x=928:16\)
\(\Rightarrow x=58\)
Vậy \(x=58.\)
Chúc bạn học tốt!
Tìm x biết:
a) \(\frac{72-x}{7}=\frac{x-40}{9}\)
b) \(\frac{37-x}{x+13}=\frac{3}{7}\)
c) \(\frac{x-1}{x+2}=\frac{x-2}{x+3}\)
a) (72-x).9=7.(x-40)
648-9x=7x-280
648+280=7x+9x (chuyển vế)
928=16x
x=928:16=58
b) (37-x).7=3.(x+13)
259-7x=3x+39
259-39=3x+7x
220=10x
x=220:10=22
c)(x-1).(x+3)=(x-2).(x+2)
x2+3x-x-3=x2+2x-2x-4
(x2-x2)+(3x-x-2x+2x)=-4+3
0+2x=-1
x=-1/2
Tìm x biết :
\(\frac{72-x}{7}=\frac{x-40}{9}\)
help me
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Bạn tham khảo nhé! Mình đã làm ở đây rồi
\(\frac{x}{5}=\frac{y}{7}=\frac{z}{4}\)và x-y+z=-10
\(\frac{x}{5}=\frac{y}{-4}=\frac{z}{-7}\)và x+y-z=-40
\(\frac{x}{2}=\frac{y}{3}=\frac{z}{7}\)và x-y+z=144
\(\frac{x}{7}=\frac{y}{8}=\frac{z}{9}\)và x+y+z=72
\(\frac{x}{6}=\frac{y}{4}=\frac{z}{3}\) và x+y-z=21
a./ \(\frac{x}{5}=\frac{y}{7}=\frac{z}{4}=\frac{x-y+z}{5-7+4}=\frac{-10}{2}=-5\)
\(\Rightarrow x=-25;y=-35;z=-20\)
b./ \(\frac{x}{5}=\frac{y}{-4}=\frac{z}{-7}=\frac{x+y-z}{5-4-\left(-7\right)}=\frac{-40}{6}=-5\)
\(\Rightarrow x=-25;y=20;z=35\)