\(2x\left(x+3\right)-\left(x-5\right)\left(7+2x\right)=44\\ \Leftrightarrow2x^2+6x-7x-2x^2+35+10x-44=0\\ \Leftrightarrow9x=9\\ \Leftrightarrow x=1\)

\(\left(2+x\right)\left(x^2-2x+4\right)-\left(3+x^2\right)\cdot x=14\\ \Leftrightarrow x^3+8-x^3-3x=14\\ \Leftrightarrow\left(x^3-x^3\right)-3x+8=14\\ \Leftrightarrow-3x+8=14\\ \Leftrightarrow-3x=6\\ \Leftrightarrow x=-2\)

\(6x^2-\left(2x+5\right)\left(3x-2\right)=7\\ \Leftrightarrow6x^2-6x^2+4x-15+10=7\\ \Leftrightarrow-11x=-3\\ \Leftrightarrow x=\dfrac{3}{7}\)

\(\left[\left(6x-39\right):7\right]\cdot4=12\\ \left(6x-39\right):7=12:4\\ \left(6x-39\right):7=3\\ 6x-39=3\cdot7\\ 6x-39=21\\ 6x=21+39\\ 6x=60\\ x=10\)

\(9\left(x+4\right)-25=5\cdot4\\ 9\left(x+4\right)-25=20\\ 9\left(x+4\right)=20+25\\ 9\left(x+4\right)=45\\ x+4=45:9\\ x+4=5\\ x=1\)

mik sửa lại ạ 😢😢

Áp dụng pytago vào \(\Delta ABC\) có :

\(BC=\sqrt{AB^2+AC^2}=\sqrt{4^2+7,5^2}=8,5\left(cm\right)\)

Áp dungj HTL:

\(AC^2=HC\cdot BC\\ \Rightarrow HC^2=\dfrac{AC^2}{BC}=\dfrac{7,5^2}{8,5}\approx6,62\left(cm\right)\\ \Rightarrow AH=\sqrt{AC^2-HC^2}=\sqrt{7,5^2-6,62^2}=3,52\left(cm\right)\)

Bài 1: 

\(a,\\ \dfrac{2}{5}-\dfrac{2}{5}x=\dfrac{2}{5}\\ \dfrac{2}{5}x=\dfrac{2}{5}-\dfrac{2}{5}\\ \dfrac{2}{5}x=0\\ x=0\)

~~~~~~~~~~~~~~~~

\(b,\\ 1\dfrac{1}{2}x-\dfrac{3}{2}=0,5\\ \dfrac{3}{2}x-\dfrac{3}{2}=\dfrac{1}{2}\\ \dfrac{3}{2}x=\dfrac{1}{2}+\dfrac{3}{2}\\ \dfrac{3}{2}x=2\\ x=2:\dfrac{3}{2}\\ x=\dfrac{4}{3}\)

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\(c,\\ \dfrac{-3}{7}-x=\dfrac{4}{5}+\left(-\dfrac{2}{3}\right)\\ \dfrac{-3}{7}-x=\dfrac{2}{15}\\ x=-\dfrac{3}{7}-\dfrac{2}{15}\\ x=-\dfrac{59}{105}\)

~~~~~~~~~~~~~~~~

\(d,\\ \dfrac{4}{7}+\dfrac{5}{9}:x=\dfrac{1}{5}\\ \dfrac{5}{9}:x=\dfrac{1}{5}-\dfrac{4}{7}\\ \dfrac{5}{9}:x=-\dfrac{13}{35}\\ x=\dfrac{5}{9}:\left(-\dfrac{13}{35}\right)\\ x=-\dfrac{175}{117}\)

\(x:8=\dfrac{7}{4}\\ x=\dfrac{7}{4}\times8\\ x=14\)

: ) quá tgian r kh sửa đc 

`@` 

\(\widehat{xOy}+\widehat{x'Oy}=180^0\) (kề bù)

\(\Rightarrow30^0+\widehat{xOy}=180^0\\ \Rightarrow\widehat{xOy}=180^0-30^0\\ \Rightarrow\widehat{xOy}=150^0\)

`@`

\(\widehat{x'Oy'}+\widehat{xOy'}=180^0\) ( kề bù )

\(\Rightarrow30^0+\widehat{x'Oy'}=180^0\\ \Rightarrow\widehat{x'Oy'}=150^0\)