Tìm x biết
\(\frac{72-x}{7}=\frac{x-40}{9}\)
(làm 2 cách
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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 \(x=58\)
\(\frac{x-40}{9}=\frac{72-x}{7}\)
=> 7 ( x - 40 ) = 9 ( 72 - x )
7 x - 280 = 648 - 9 x
7 x + 9 x = 648 + 280
16 x = 928
x = 58
\(\frac{x-40}{9}=\frac{72-x}{7}\)
\(\Rightarrow\left(x-40\right).7=9.\left(72-x\right)\)
\(\Rightarrow648-9x=7x-280\)
\(\Rightarrow\left(-280\right)-648=\left(-9x\right)-7x\)
\(\Rightarrow-928=-16x\)
\(\Rightarrow x=58\)
=> (72 - x) . 9 = (x - 40) . 7
=> 648 - 9x = 7x - 280
=> 648 + 280 = 7x + 9x
=> 928 = 16x
=> x = 58
\(\frac{x-2}{12}+\frac{x-2}{20}+\frac{x-2}{30}+\frac{x-2}{42}+\frac{x-2}{56}+\frac{x-2}{72}=\frac{16}{9}\)
\(\left(x-2\right)\cdot\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\right)=\frac{16}{9}\)
\(\left(x-2\right)\cdot\left(\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\right)=\frac{16}{9}\)
\(\left(x-2\right)\cdot\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)=\frac{16}{9}\)
\(\left(x-2\right)\cdot\left(\frac{1}{3}-\frac{1}{9}\right)=\frac{16}{9}\)
\(\left(x-2\right)\cdot\left(\frac{3}{9}-\frac{1}{9}\right)=\frac{16}{9}\)
\(\left(x-2\right)\cdot\frac{2}{9}=\frac{16}{9}\)
\(x-2=\frac{16}{9}:\frac{2}{9}\)
\(x-2=\frac{16}{9}\cdot\frac{9}{2}\)
\(x-2=8\)
\(x=8+2\)
\(x=10\)
Vậy \(x=10\)
\(\left(x-2\right)\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\right)=\)\(=\frac{16}{9}\)
\(\left(x-2\right)\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}+\frac{1}{8}-\frac{1}{9}\right)=\frac{16}{9}\)
\(\left(x-2\right)\left(\frac{1}{3}-\frac{1}{9}\right)=\frac{16}{9}\)
\(\left(x-2\right)\left(\frac{2}{9}\right)=\frac{16}{9}\)
2(x-2)=16
x-2=8
x=10
=> x-2.(1/12+1/20+1/30+1/42+1/56+1/72)=16/9
Đặt : Sáng = 1/12+1/20+1/30+1/42+1/56+1/72
=> Sáng = 1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9
=> Sáng = 1.(1/3-1/4+1/4-1/5+...+1/8-1/9
=> Sáng = 91.(1/3-1/9)
=> Sáng = 2/9
Thay Sáng vô biểu thức 1/12+1/20+1/30+1/42+1/56+1/72
Ta được :
x-2.2/9=16/9
giờ thì tự làm nha
=> x-2.(1/12+1/20+1/30+1/42+1/56+1/72)=16/9
Đặt : Sáng = 1/12+1/20+1/30+1/42+1/56+1/72
=> Sáng = 1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9
=> Sáng = 1.(1/3-1/4+1/4-1/5+...+1/8-1/9
=> Sáng = 91.(1/3-1/9)
=> Sáng = 2/9
Thay Sáng vô biểu thức 1/12+1/20+1/30+1/42+1/56+1/72
Ta được :
x-2.2/9=16/9
giờ thì tự làm nha
Ai k mk mk k lại
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.
\(\frac{72-9}{7}=\frac{x-40}{9}\)
\(\frac{x-40}{9}=\frac{63}{7}\)
\(x-40=\frac{63}{7}\times9\)
\(x-40=81\)
\(x=81+40\)
\(x=121\)
\(\text{Cách 1:}\)
\(\frac{72-x}{7}=\frac{x-40}{9}\)
=> \(\left(72-x\right)\cdot9=\left(x-40\right)\cdot7\)
=> \(648-9x=7x-280\)
=> \(648+280=7x+9x\)
=> \(16x=928\)=> \(x=928:16=58\)
\(\text{Cách 2:}\)
\(\text{Áp dụng tính chất dãy tỉ số bằng nhau, ta có:}\)
\(\frac{72-x}{7}=\frac{x-40}{9}=\frac{72-x+x-40}{7+9}=\frac{32}{16}=2\)
\(\text{Ta suy ra:}\)
\(\frac{72-x}{7}=2\)=> \(72-x=2\cdot7=14\)=> \(x=72-14=58\)
\(\text{Vậy }x=58\)