các bạn ơi giúp mình với

2⁵ = 2².2³ < 2².3² = 6² = 2².3² < 3².3² < 3⁵

Vậy 2⁵ < 6² < 3⁵ (đpcm)

Tỉ số của tam giác CMB trên tam giác ABC là: \(\dfrac{CMB}{ABC}=\dfrac{MB}{AB}=\dfrac{1}{2}\)

Diện tích tam giác CMB là: \(\dfrac{1}{2}\cdot60=30\)

Tỉ số của CD trên BC là : \(\dfrac{CD}{BC}=\dfrac{CMD}{CMB}=\dfrac{10}{30}=\dfrac{1}{3}\)

Đ/S 1/3

\(a.\left[x\cdot\left(x+1\right)\right]:2=136\\ x\cdot\left(x+1\right)=136\cdot2\\ x\cdot\left(x+1\right)=2\cdot2\cdot2\cdot17\cdot2\\ x\cdot\left(x+1\right)=\left(2\cdot2\cdot2\cdot2\right)\cdot17\\ x\cdot\left(x+1\right)=16\cdot17\\ =>x=16\\ b.\left[x\cdot\left(x+1\right)\right]:2=300\\ x\cdot\left(x+1\right)=3\cdot2\cdot2\cdot5\cdot5\cdot2\\ x\cdot\left(x+1\right)=\left(3\cdot2\cdot2\cdot2\right)\cdot\left(5\cdot5\right)\\ x\cdot\left(x+1\right)=24\cdot25\\ =>x=24\\ c.\left[x\cdot\left(x+1\right)\right]:2=561\\ x\cdot\left(x+1\right)=2\cdot3\cdot11\cdot17\\ x\cdot\left(x+1\right)=\left(17\cdot2\right)\cdot\left(3\cdot11\right)\\ x\cdot\left(x+1\right)=34\cdot33\\ =>x=33\)

\(x\cdot\left(x+1\right):2=153\)

\(\Rightarrow x\left(x+1\right)=153\cdot2\)

\(\Rightarrow x^2+x=306\)

\(\Rightarrow x^2+x-306=0\)

\(\Rightarrow x^2+18x-17x-306=0\)

\(\Rightarrow x\left(x+18\right)-17\left(x+18\right)=0\)

\(\Rightarrow\left(x-17\right)\left(x+18\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=17\\x=-18\end{matrix}\right.\)

\(x\left(x+1\right)=156\)

\(\Rightarrow x^2+x=156\)

\(\Rightarrow x^2+x-156=0\)

\(\Rightarrow x^2+13x-12x-156=0\)

\(\Rightarrow x\left(x+13\right)-12\left(x+13\right)=0\)

\(\Rightarrow\left(x+13\right)\left(x-12\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=12\\x=-13\end{matrix}\right.\)

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\(x\left(x+1\right)=342\)

\(\Rightarrow x^2+x=342\)

\(\Rightarrow x^2+x-342=0\)

\(\Rightarrow x^2+19x-18x-342=0\)

\(\Rightarrow x\left(x+19\right)-18\left(x+19\right)=0\)

\(\Rightarrow\left(x+19\right)\left(x-18\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-19\\x=18\end{matrix}\right.\)

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\(x\left(x+1\right)=650\)

\(\Rightarrow x^2+x=650\)

\(\Rightarrow x^2-x+650=0\)

\(\Rightarrow x^2+26x-25x-650=0\)

\(\Rightarrow x\left(x+26\right)-25\left(x+26\right)=0\)

\(\Rightarrow\left(x+26\right)\left(x-25\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-26\\x=25\end{matrix}\right.\)

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\(x\left(x+1\right)=380\)

\(\Rightarrow x^2+x=380\)

\(\Rightarrow x^2+x-380=0\)

\(\Rightarrow x^2+20x-19x-380=0\)

\(\Rightarrow x\left(x+20\right)-19\left(x+20\right)=0\)

\(\Rightarrow\left(x+20\right)\left(x-19\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-20\\x=19\end{matrix}\right.\)

a, \(x\).(\(x\) + 1) = 156

156 = 2².3.13 = 12.13

Vậy \(x\).(\(x\) + 1) = 12.13

Vậy \(x\) = 12

b, \(x.\)(\(x\) + 1) =  342 

    342 = 2.3².19 = 18.19

    \(x\).(\(x+1\)) = 18.19

    \(x\) =   18

c, \(x\).(\(x\) + 1) = 650 

    650 = 2.5².13 = 25.26

    \(x\).(\(x\) +1) = 25.26

    \(x\) = 25

d, \(x\).(\(x\) +1) = 380 

    380 = 2².5.19 = 19.20

      \(x\).(\(x\) + 1) = 19.20

      \(x\)    = 19

\(\dfrac{15}{90}\) = \(\dfrac{1}{6}\) = \(\dfrac{1.5}{6.5}\) = \(\dfrac{5}{30}\)

\(\dfrac{120}{600}\) = \(\dfrac{1}{5}\) = \(\dfrac{1.6}{5.6}\) = \(\dfrac{6}{30}\)

\(\dfrac{75}{150}\) = \(\dfrac{1}{2}\) = \(\dfrac{1.15}{2.15}\) = \(\dfrac{15}{30}\)

Ta có:

\(\dfrac{15}{90}=\dfrac{1}{6};\dfrac{120}{600}=\dfrac{1}{5};\dfrac{75}{150}=\dfrac{1}{2}\)

\(\Rightarrow MSC\) là: \(60\)

\(\dfrac{1}{6}=\dfrac{1.10}{6.10}=\dfrac{10}{60}\)

\(\dfrac{1}{5}=\dfrac{1.12}{5.12}=\dfrac{12}{60}\)

\(\dfrac{1}{2}=\dfrac{1.30}{2.30}=\dfrac{30}{60}\)

Ta có:

\(\dfrac{54}{90}=\dfrac{3}{5}\)

\(\dfrac{180}{288}=\dfrac{5}{8}\)

\(\dfrac{60}{135}=\dfrac{4}{9}\)

Mẫu số chung là: 360

\(\dfrac{3}{5}=\dfrac{3.72}{5.72}=\dfrac{216}{360}\)

\(\dfrac{5}{8}=\dfrac{5.45}{8.45}=\dfrac{225}{360}\)

\(\dfrac{4}{9}=\dfrac{4.40}{9.40}=\dfrac{160}{360}\)

giải giúp mình với

Diện tích tam giác $ABC$ là $:$

$S_{\Delta ABC}=\frac{a\times h}{2}=\frac{20\times 12}{2}=120\left(c{m}^2\right)$

$b,$ Nối $A$ với $M.$ Theo giả thiết ta có $:$

$S_{\Delta ABM}=\frac{1}{2}{S}_{\Delta ABC}=\frac{1}{2}.120=60c{m}^2$

$S_{\Delta MBN}=\frac{1}{2}{S}_{\Delta ABM}=60\frac{.1}{2}=30c{m}^2$

Từ đó cũng có $:$ $S_{\Delta MBN}=S_{\Delta PMC}$

$\Rightarrow$ $S_{\Delta MNP}=120-3\times 30=30\left(c{m}^2\right)$

Ta có:

\(\dfrac{20}{45}=\dfrac{5}{9}\)

\(\dfrac{14}{35}=\dfrac{2}{5}\)

\(\dfrac{32}{44}=\dfrac{8}{11}\) 

Mẫu số chung là:  495

\(\dfrac{5}{9}=\dfrac{5\cdot55}{9\cdot55}=\dfrac{275}{495}\)

\(\dfrac{2}{5}=\dfrac{2\cdot99}{5\cdot99}=\dfrac{198}{495}\)

\(\dfrac{8}{11}=\dfrac{8\cdot45}{11\cdot45}=\dfrac{360}{495}\)

 \(\dfrac{20}{45}\) = \(\dfrac{4}{9}\) = \(\dfrac{4.5.11}{9.5.11}\) = \(\dfrac{220}{495}\)

 \(\dfrac{14}{35}\) = \(\dfrac{2}{5}\) = \(\dfrac{2.9.11}{9.5.11}\) = \(\dfrac{198}{495}\)

  \(\dfrac{32}{44}\) = \(\dfrac{8}{11}\) = \(\dfrac{8.9.5}{9.5.11}\)= \(\dfrac{360}{495}\)