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

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

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

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

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

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

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

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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

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

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

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

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

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

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

Ta có:

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

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

\(\dfrac{60}{135}=\dfrac{4}{9}\) 

Có mẫu số chung là 360

\(\dfrac{3}{5}=\dfrac{3\cdot72}{5\cdot72}=\dfrac{216}{360}\)

\(\dfrac{5}{8}=\dfrac{5\cdot45}{8\cdot45}=\dfrac{225}{360}\)

\(\dfrac{4}{9}=\dfrac{4\cdot40}{9\cdot40}=\dfrac{160}{360}\)

1. Gọi số cần tìm là: abcdef

Ta có: abcde4

         4abcde = abcde4 . 4

         4 . 100000 + abcde = (abcde . 10 + 4)

Đạt abcde = x, ta có:

         400000 + x = (x . 10 + 4) . 4

         400000 + x = x . 10 . 4 + 4 . 4

         400000 + x = x . 40 + 16

         400000 - 16 = x . 40 - x 

               399984 = x . 39

                x = 399984 : 39

                x =          102564

Vậy số cần tìm là: 102564

có thể lập được 27 số

\(MSC\left(150;50;75\right)=150\)

\(\dfrac{17}{150}\)

\(\dfrac{7}{50}=\dfrac{14}{150}\)

\(\dfrac{4}{75}=\dfrac{16}{150}\)

Đính chính \(\dfrac{7}{50}=\dfrac{21}{150}\)

\(8\left(x+9\right)+4\left(6-x\right)=98\)

\(\Rightarrow8x+72+24-4x=98\)

\(\Rightarrow4x+98=98\)

\(\Rightarrow4x=98-98\)

\(\Rightarrow4x=0\Rightarrow x=0\)

8(x + 9) + 4(6 - x) + 2 = 98
8x + 72 + 4(6 - x) + 2 = 98
8x + 72 + 4(-x + 6) + 2 = 98
8x + 72-4x + 24 + 2 = 98
8x + 98 -4x = 98
4x + 98 = 98
4x + 98-98 = 98-98
4x = 98 - 98
4x = 0
4x/4 = 0/4
x = 0/4
x = 0