`a)[2x-3]/4-[4x-5]/3=[5-x]/6`

`<=>3(2x-3)-4(4x-5)=2(5-x)`

`<=>6x-9-16x+20=10-2x`

`<=>8x=1`

`<=>x=1/8`

(`b->c` mk đã làm r).

\(\dfrac{4}{x+3}+\dfrac{x+7}{x^2-9}\)

\(=\dfrac{4\left(x-3\right)+x+7}{x^2-9}\)

\(=\dfrac{4x-12+x+7}{x^2-9}\)

\(=\dfrac{5x-5}{x^2-9}\)

Bài 1:
\(a,\left(x+4\right)\left(x+3\right)-7x=x^2+4x+3x+12-7x=x^2+12\\ b,\left(x+4\right)^2+x-16=x^2+8x+16+x-16=x^2+9x\\ c,\dfrac{4}{x+3}+\dfrac{x+7}{x^2-9}=\dfrac{4\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\dfrac{x+7}{\left(x+3\right)\left(x-3\right)}=\dfrac{4x-12+x+7}{\left(x+3\right)\left(x-3\right)}=\dfrac{5x-5}{\left(x+3\right)\left(x-3\right)}\)

Bài 2:
\(7a-7b=7\left(a-b\right)\\ b,x^2-8x+16=\left(x-4\right)^2\\ c,ax-ay+3x-3y=a\left(x-y\right)+3\left(x-y\right)=\left(a+3\right)\left(x-y\right)\\ d,x^2+6x+9-y^2=\left(x+3\right)^2-y^2=\left(x-y+3\right)\left(x+y+3\right)\)

a/\(-5,94-13,78=-19,72\)
b/\(\dfrac{3}{4}-\dfrac{5}{4}:\dfrac{-10}{16}\)
\(=\dfrac{3}{4}+\dfrac{5}{4}\cdot\dfrac{8}{5}\)
\(=\dfrac{3}{4}+\dfrac{8}{4}\)
\(=\dfrac{11}{4}\)
c/\(\dfrac{12}{17}\cdot\dfrac{5}{9}-\dfrac{5}{9}\cdot\dfrac{29}{17}+1\dfrac{5}{9}\)
\(=\left(\dfrac{12}{17}-\dfrac{29}{17}\right)\dfrac{5}{9}+1\dfrac{5}{9}\)
\(=-1\cdot\dfrac{5}{9}+1\dfrac{5}{9}\)
\(=\dfrac{-5}{9}+1\dfrac{5}{9}\)
\(=1\)

Chọn A

12m2 6dm2 = 12,06 m2

7ha 16a=7,16 ha

4m2 32dm2 = 0,0432 a

23 ha 512 m2 = 23,0512 ha

5,6 km2 = 560 ha

21 ha = 0,21 km2

7km2 241 m2 = 700,0241 ha

7 820 000 m2 = 7,82 km2

12m2 6dm2= dm2

12m2 6dm2 = 12,06 m2

7ha 16a=7,16 ha

4m2 32dm2 = 0,0432 a

23 ha 512 m2 = 23,0512 ha

5,6 km2 = 560 ha

21 ha = 0,21 km2

7km2 241 m2 = 700,0241 ha

7 820 000 m2 = 7,82 km2

a) \(\dfrac{29}{60}\)

b)\(1\dfrac{23}{25}\)

c) \(2\dfrac{26}{35}\)

Đề câu a sai rồi.

`a,(x-3)^{x+8}=4(x-3)^{x+6}`

`=>(x-3)^{x+6}[(x-3)^2-4]=0`

`=>` $\left[ \begin{array}{l}x-3=0\\(x-3)^2=4\end{array} \right.$

`=>` $\left[ \begin{array}{l}x=3\\x-3=2\\x-3=-2\end{array} \right.$

`=>` $\left[ \begin{array}{l}x=3\\x=5\\x=1\end{array} \right.$

Vậy x=1 hoặc x=3 hoặc x=5.

`b,(x-3)^{x+10}=9(x-3)^{x+8}`

`=>(x-3)^{x+8}[(x-3)^2-9]=0`

`=>` $\left[ \begin{array}{l}x-3=0\\(x-3)^2=9\end{array} \right.$

`=>` $\left[ \begin{array}{l}x=3\\x-3=3\\x=-3=-3\end{array} \right.$

`=>` $\left[ \begin{array}{l}x=3\\x=6\\x=0\end{array} \right.$

Vậy x=0 hoặc x=3 hoặc x=6