a, Áp dụng các t/c các số tận cùng là 1 và 6khi tăng bậc số tận cùng vẫn là 6 và 6.
2²⁰¹⁵=2.2²⁰¹⁴=2.4¹⁰⁰⁷=2.4.4¹⁰⁰⁶=8.16⁵⁰³=8.(...6)=(...8)
3²⁰¹⁴=9¹⁰⁰⁷=9.9¹⁰⁰⁶=9.81⁵⁰³=9.(...1)=(...9)
=2²⁰¹⁵ + 3²⁰¹⁴ =(...8)+(...9)=(...7)
b, 17²⁰²³≡7²⁰²³=7.7²⁰²²=7.49¹⁰¹¹=7.49.49¹⁰¹⁰=7.49.2401⁵⁰⁵=(...3)

d) \(\left(x+2\right)\left(x^2-2x+4\right)\)

\(=\left(x+2\right)\left(x^2-2\cdot x+2^2\right)\)

\(=x^3+2^3\)

\(=x^3+8\)

e) \(\left(\dfrac{1}{4}-\dfrac{x}{5}\right)\left(\dfrac{x^2}{25}+\dfrac{x}{20}+\dfrac{1}{16}\right)\)

\(=\left(\dfrac{1}{4}-\dfrac{1}{5}x\right)\left(\dfrac{1}{25}x^2+\dfrac{1}{5}x\cdot\dfrac{1}{4}+\dfrac{1}{16}\right)\)

\(=\left(\dfrac{1}{4}-\dfrac{1}{5}x\right)\left[\left(\dfrac{1}{5}x\right)^2+\dfrac{1}{5}x\cdot\dfrac{1}{4}+\left(\dfrac{1}{4}\right)^2\right]\)

\(=\left(\dfrac{1}{4}\right)^3-\left(\dfrac{1}{5}x\right)^3\)

\(=\dfrac{1}{64}-\dfrac{1}{125}x^3\)

\(=\dfrac{1}{64}-\dfrac{x^3}{125}\)

d: (x+2)(x^2-2x+4)

=(x+2)(x^2-x*2+2^2)

=x^3+8

e: (1/4-x/5)(1/16+x/20+x^2/25)

=(1/4-x/5)[(1/4)^2+1/4*x/5+(x/5)^2]

=1/64-x^3/125

Số tiền 1 ngày phải dùng là:

16000*200=3200000

25600000/3200000=8

=>Có thể mua được số gạo để cho 200 hộ dùng được trong 8 ngày

Số tiền 1 ngày phải dùng là:

16000*200=3200000

vậy 25600000/3200000=8

=>Có thể mua được số gạo để cho 200 hộ dùng được trong 8 ngày

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

\(9x^2-7\)

\(=\left(3x\right)^2-\left(\sqrt{7}\right)^2\)

\(=\left(3x+\sqrt{7}\right)\left(3x-\sqrt{7}\right)\)

he hadn't drunk alcohol, he would have passed the contest.

you change your attitude and become friendlier, you'll lose your job.

somebody waters these flowers, they will die.

=>4x+2=2x+2

=>2x=0

=>x=0

\(7^{4x}+2=49^x+1\)

\(\Leftrightarrow4x+2=2\left(x+1\right)\)

\(\Leftrightarrow4x+2=2x+2\)

\(\Leftrightarrow4x-2x=2-2\)

\(\Leftrightarrow2x=0\)

\(\Leftrightarrow x=0\)

x= 0 nha

\(x^{23}=x^6\)

\(\Rightarrow x^{23}-x^6=0\)

\(\Rightarrow x^6\cdot\left(x^{17}-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x^6=0\\x^{17}-1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0^6\\x^{17}=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=1^{17}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

=>x^6(x^17-1)=0

=>x=0 hoặc x=1

\(\left(12-6\sqrt{3}\right)\sqrt{\dfrac{3}{14-8\sqrt{3}}}-3\sqrt{2\left(1-\sqrt{-2\sqrt{3}+4}\right)+2\sqrt{4+2\sqrt{3}}}\)

\(=\left(12-6\sqrt{3}\right).\dfrac{\sqrt{3}}{\sqrt{14-8\sqrt{3}}}-3\sqrt{2\left(1-\sqrt{\left(\sqrt{3}-1\right)^2}\right)+2\sqrt{\left(\sqrt{3}+1\right)^2}}\)

\(=\left(12-6\sqrt{3}\right).\dfrac{\sqrt{3}}{\sqrt{14+2\sqrt{48}}}-3\sqrt{2\left(1-\left|\sqrt{3}-1\right|\right)+2\left|\sqrt{3}+1\right|}\)

\(=\left(12-6\sqrt{3}\right).\dfrac{\sqrt{3}}{\sqrt{\left(\sqrt{8}+\sqrt{6}\right)^2}}-3\sqrt{2\left(1-\sqrt{3}+1\right)+2\sqrt{3}+2}\)

\(=\dfrac{12\sqrt{3}-18}{\left|\sqrt{8}+\sqrt{6}\right|}-3\sqrt{2\left(2-\sqrt{3}\right)+2\sqrt{3}+2}\)

\(=\dfrac{12\sqrt{3}-18}{\sqrt{8}+\sqrt{6}}-3\sqrt{4-2\sqrt{3}+2\sqrt{3}+2}\)

\(=\dfrac{12\sqrt{3}-18}{\sqrt{8}+\sqrt{6}}-3\sqrt{6}\\ =\dfrac{12\sqrt{3}-18-3\sqrt{6}\left(\sqrt{8}+\sqrt{6}\right)}{\sqrt{8}+\sqrt{6}}\\ =\dfrac{12\sqrt{3}-18-18-12\sqrt{3}}{\sqrt{8}+\sqrt{6}}\\ =\dfrac{-36}{\sqrt{8}+\sqrt{6}}\)

\(=\left(12-6\sqrt{3}\right)\cdot\sqrt{\dfrac{21+12\sqrt{3}}{2}}-3\cdot\sqrt{2\cdot\left(1-\sqrt{1-\sqrt{3}+1}\right)+2\sqrt{3}+2}\)

\(=\left(12-6\sqrt{3}\right)\cdot\dfrac{2\sqrt{3}+3}{\sqrt{2}}-3\cdot\sqrt{2-\sqrt{2}\left(\sqrt{4-2\sqrt{3}}\right)+2\left(\sqrt{3}+1\right)}\)

\(=3\sqrt{6}-3\cdot\sqrt{2-\sqrt{2}\left(\sqrt{3}-1\right)+2\sqrt{3}+2}\)

\(=3\sqrt{6}-3\cdot\sqrt{4+2\sqrt{3}-\sqrt{6}+\sqrt{2}}\)

a) \(\left(\dfrac{x^2}{2}+y^2\right)^2\)

\(=\left(\dfrac{1}{2}x^2+y^2\right)^2\)

\(=\left(\dfrac{1}{2}x^2\right)^2+2\cdot\dfrac{1}{2}x^2\cdot y^2+\left(y^2\right)^2\)

\(=\dfrac{1}{4}x^4+x^2y^2+y^4\)

b) \(\left(\dfrac{4}{5}x^2-\dfrac{2}{3}y\right)^2\)

\(=\left(\dfrac{4}{5}x^2\right)^2-2\cdot\dfrac{4}{5}x^2\cdot\dfrac{2}{3}y+\left(\dfrac{2}{3}y\right)^2\)

\(=\dfrac{16}{25}x^4-\dfrac{16}{15}x^2y+\dfrac{4}{9}y^2\)

c) \(\left(2x+\dfrac{1}{2}\right)\left(2x-\dfrac{1}{2}\right)\)

\(=\left(2x\right)^2-\left(\dfrac{1}{2}\right)^2\)

\(=4x^2-\dfrac{1}{4}\)

a: (1/2x^2+y^2)^2

=(1/2x^2)^2+2*1/2x^2*y^2+y^4

=1/4x^4+x^2y^2+y^4

b: (4/5x^2-2/3y)^2

=(4/5x^2)^2-2*4/5x^2*2/3y+4/9y^2

=16/25x^4-16/15x^2y+4/9y^2

c: =(2x)^2-(1/2)^2

=4x^2-1/4

1. Unless he hurries, he'll be late for school.

2. If it stops raining, we'll go out.

3. If he didn't study harder, he would fail the final exam.

unless he hurries, he will be late for school

if it stops raining, we will go out

if he didn't study harder, he would fail the final exam