\(76^2-24^2=\left(76-24\right)\left(76+24\right)=52.100=5200\)

\(24^2+48.76+76^2=24^2+2.24.76+76^2=\left(24+76\right)^2=100^2=10000\)

\(23,18.26+44.23,18=23,18\left(26+44\right)=23,18.70=1622,6\)

\(53.47=\left(50+3\right)\left(50-3\right)=50^2-3^2=2500-9=2491\)

Olm chào em, bài này thì em làm như sau em nhé:

                              

b c b c d b d b b d

1 B

2 C

3 A

4 C

5 A

6 B

7 D

8 B

9 B

10 C

Bài 4:

\(a,\Rightarrow5⋮x\Rightarrow x\inƯ\left(5\right)=\left\{1;5\right\}\\ b,\Rightarrow x-2+7⋮x-2\\ \Rightarrow x-2\inƯ\left(7\right)=\left\{1;7\right\}\\ \Rightarrow x\in\left\{3;9\right\}\\ c,\Rightarrow3\left(x+1\right)+4⋮x+1\\ \Rightarrow x+1\inƯ\left(4\right)=\left\{1;2;4\right\}\\ \Rightarrow x\in\left\{0;1;3\right\}\\ d,\Rightarrow10x+6⋮2x-1\\ \Rightarrow5\left(2x-1\right)+11⋮2x-1\\ \Rightarrow2x-1\inƯ\left(11\right)=\left\{1;11\right\}\\ \Rightarrow x\in\left\{1;6\right\}\\ e,\Rightarrow x\left(x+3\right)+11⋮x+3\\ \Rightarrow x+3\inƯ\left(11\right)=\left\{1;11\right\}\\ \Rightarrow x=8\left(x\in N\right)\\ f,\Rightarrow x\left(x+3\right)+2\left(x+3\right)+5⋮x+3\\ \Rightarrow x+3\inƯ\left(5\right)=\left\{1;5\right\}\\ \Rightarrow x=2\left(x\in N\right)\)

a) x=16

b)x=5

c)x=1

Câu cuối ko bt tính có đúng hông?! Nếu sai choa mk "Sỏ ri"nhá!!!

Bài 1

a/ 

\(\frac{4}{5}=\frac{x}{20}\)

Quy đồng, ta có : \(\frac{4}{5}=\frac{16}{20}\Rightarrow x-16\)

b/ \(\frac{x}{7}=\frac{30}{42}\)

Rút gọn, ta được : \(\frac{30}{42}=\frac{5}{7}\Rightarrow x=5\)

c/ \(\frac{18}{5}\times x=\frac{9}{7}:\frac{5}{14}\)

     \(\frac{18}{5}\times x=\frac{18}{5}\)

      \(\Rightarrow x=1\)

c, 

\(\frac{18}{5}\cdot x=\frac{9}{7}\div\frac{5}{14}\)

\(\Rightarrow\frac{18}{5}\cdot x=\frac{18}{5}\)

\(\Rightarrow x=1\)

vậy_

bn chụp màn hình sao vậy? k bt giống bn k alt+ctrl+a

Bài 5: 

a: \(x\left(x-1\right)-x^2+4x=-3\)

\(\Leftrightarrow x^2-x-x^2+4x=-3\)

hay x=-1

b: \(6x^2-\left(2x+5\right)\left(3x-2\right)=7\)

\(\Leftrightarrow6x^2-6x^2+4x-15x+10=7\)

\(\Leftrightarrow-11x=-3\)

hay \(x=\dfrac{3}{11}\)

Bài 5: 

a: \(x\left(x-1\right)-x^2+4x=-3\)

\(\Leftrightarrow x^2-x-x^2+4x=-3\)

hay x=-1

i: \(x^2-9x+8=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=8\end{matrix}\right.\)