Lời giải:

$2^{56}=(2^{28})^2> (2^3)^2=8^2> 5^2$

2⁵⁶ và 5² 

2⁵⁶ = (2³)².2⁵⁰ = 2⁵⁰.8² >5² 

vậy 2⁵⁶ > 5² 

d.(3/4 x 8/9) x [(15/16 x 24/25) x35/36]

=2/3 x 7/8= 7/12

3/4 x 8/9 x 15/16 x 24/25 x 35/36

=(3/4 x 8/9) x 15/16 x (24/25 x 35/36)

=2/3 x 15/16 x 14/15

=2/3 x (15/16 x 14/15)

=2/3 x 7/8

=7/12

?

\(A\left(x\right)=4x^2-2x-8+5x^3-7x^2+1\)

\(A\left(x\right)=5x^3-3x^2-2x-7\)

\(B\left(x\right)=-3x^3+4x^2+9+x-2x-2x^3\)

\(B\left(x\right)=-5x^3+4x^2-x+9\)

+) \(A\left(x\right)=4x^2-2x-8+5x^3-7x^2+1\)

\(\Leftrightarrow A\left(x\right)=-3x^2-2x-7+5x^3\)

+) \(B\left(x\right)=-3x^3+4x^2+9+x-2x-2x^3\)

\(\Leftrightarrow B\left(x\right)=-5x^3+4x^2-x+9\)

+) \(M\left(x\right)=A\left(x\right)+B\left(x\right)\)

\(\Leftrightarrow M\left(x\right)=-3x^2-2x-7+5x^3-5x^3+4x^2-x+9\)

\(\Leftrightarrow M\left(x\right)=x^2-3x+2\)

+) \(N\left(x\right)=A\left(x\right)-B\left(x\right)\)

\(\Leftrightarrow N\left(x\right)=-3x^2-2x-7+5x^3+5x^3-4x^2+x-9\)

\(\Leftrightarrow N\left(x\right)=-7x^2+10x^3-x-16\)

Sửa đề:

\(\dfrac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)

\(=\dfrac{\left(2^2\right)^6.\left(3^2\right)^5+\left(2.3\right)^9.120}{\left(2^3\right)^4.3^{12}-\left(2.3\right)^{11}}\)

\(=\dfrac{2^{12}.3^{10}+2^9.3^9.120}{2^{12}.3^{12}-2^{11}.3^{11}}\)

\(=\dfrac{2^9.3^9\left(2^3.3+120\right)}{2^{11}.3^{11}\left(2.3-1\right)}\)

\(=\dfrac{2^9.3^9.144}{2^{11}.3^{11}.5}=\dfrac{144}{180}=\dfrac{4}{5}\)

a) Vì tia Oz là tia phân giác của \(\widehat{xOy}\)

\(\Rightarrow\widehat{xOz}=\widehat{zOy}=\dfrac{1}{2}.\widehat{xOy}=\dfrac{1}{2}.80^o=40^o\)

b) Vì Om là tia phân giác của \(\widehat{xOz}\)

\(\Rightarrow\widehat{mOz}=\dfrac{1}{2}.\widehat{xOz}=\dfrac{1}{2}.40^o=20^o\)

Vì  On là tia phân giác của \(\widehat{zOy}\)

\(\Rightarrow\widehat{zOn}=\dfrac{1}{2}.\widehat{zOy}=\dfrac{1}{2}.40^o=20^0\)

+) \(\widehat{mOn}=\widehat{mOz}+\widehat{zOn}=20^o+20^o=40^o\)

Tìm một số biết rầng lớp nghìn gấp 3 lần đơn vị, lớp triệu gấp 2 lần lớp nghìn

a) Vì \(\widehat{xOy}+\widehat{yOz}=180^o\) ( 2 góc kề bù )

\(\Rightarrow\widehat{yOz}=180^o-\widehat{xOy}=180^o-100^o=80^o\)

b) Vì Om là tia phân giác của \(\widehat{xOy}\)

\(\Rightarrow\widehat{xOm}=\widehat{mOy}=\dfrac{1}{2}\widehat{xOy}=\dfrac{1}{2}.100^o=50^o\)

Vì On là tia phân giác của \(\widehat{yOz}\)

\(\Rightarrow\widehat{yOn}=\widehat{nOz}=\dfrac{1}{2}.\widehat{yOz}=\dfrac{1}{2}.80^o=40^o\)

+) \(\widehat{mOn}=\widehat{mOy}+\widehat{yOn}=50^o+40^o=90^o\)

\(\dfrac{45^0.5^{20}}{75^{15}}\)

\(=\dfrac{1.5^{20}}{\left(5^2.3\right)^{15}}\)

\(=\dfrac{5^{20}}{5^{30}.3^{15}}\)

\(=\dfrac{1}{5^{10}.3^{15}}\)