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\(\frac{40-x}{9}=\frac{28}{36}\)
=> \(\frac{40-x}{9}=\frac{7}{9}\)
\(=>40-x=7\)
\(x=40-7\)
\(x=33\). Vậy x=33
mk khẳng định chắc chắn là 33
Vì \(\frac{5}{7}=\frac{x}{63}\)nên 5.63 = 7.x
=> \(\frac{5.63}{7}\)= 45
Vậy x = 45
\(\frac{5}{7}=\frac{x}{63}\)
=> 63 . 5 = 7.x
=> 315 = 7x
=> x = \(\frac{315}{7}=45\)
Lớp 4?
a) \(\dfrac{x+1}{4}=\dfrac{36}{x+1}\)
\(\Rightarrow\left(x+1\right)^2=144\)
\(\Rightarrow\left[{}\begin{matrix}x+1=12\\x+1=-12\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=11\\x=-13\end{matrix}\right.\)
Vậy: \(x\in\left\{11;-13\right\}\)
b) \(\dfrac{x}{7}=\dfrac{55x-4}{28}\)
\(\Rightarrow4x=55x-4\)
\(\Rightarrow-51x=-4\)
\(\Rightarrow x=\dfrac{4}{51}\)
Vậy: \(x=\dfrac{4}{51}\)
a) \(\dfrac{x + 1}{4} = \dfrac{36}{x + 1} \)
\(\Rightarrow\) \(( x + 1 )( x + 1 ) = 36 . 4 \)
\(\Rightarrow ( x + 1 )^2 = 144 \)
\(\Rightarrow ( x + 1 )^2 = 12^2 = ( -12 )^2 \)
\(\Rightarrow\) \(x + 1 ∈ \) { \(12 ; -12 \) }
\(\Rightarrow \) \(x \) \(∈ \) { \(11 ; -13 \) }
Vậy \(x ∈ \) { \(11 ; -13 \) }
\(\Rightarrow28\left(x-9\right)=20.35\Rightarrow28x-252=700\Rightarrow28x=952\Rightarrow x=34\)
\(\frac{x-9}{35}=\frac{20}{28}=\frac{5}{7}=\frac{25}{35}\)
=>x-9=25
=>x=25+9
=>x=34
vậy x=34
a) $\frac{2}{3} = \frac{{2 \times 6}}{{3 \times 6}} = \frac{{12}}{{18}}$
Ta có $\frac{{12}}{{18}} > \frac{{11}}{{18}}$ nên $\frac{2}{3} > \frac{{11}}{{18}}$
b) $\frac{{36}}{{63}} = \frac{{36:9}}{{63:9}} = \frac{4}{7}$
Ta có $\frac{4}{7} < \frac{5}{7}$ nên $\frac{{36}}{{63}}$ < $\frac{5}{7}$
c)
$\frac{{55}}{{110}} = \frac{{55:55}}{{110:55}} = \frac{1}{2}$ ; $\frac{4}{8} = \frac{1}{2}$
Vậy $\frac{{55}}{{110}}$ = $\frac{4}{8}$
\(\frac{28}{x+4}=\frac{36}{63}=\frac{4}{7}\Leftrightarrow\frac{7.28}{7\left(x+4\right)}=\frac{4\left(x+4\right)}{7\left(x+4\right)}\)
=> 4(x+4) = 7.28
=> 4x + 16 = 196
=> 4x = 180
=> x = 4 5