b, \(4x^2-7x+3=4x^2-4x-3x+3\)

\(=4x\left(x-1\right)-3\left(x-1\right)\)

\(=\left(4x-3\right)\left(x-1\right)\)

c, \(\left(3x-1\right)^2-\left(2x-3\right)^2\)

\(=\left(3x-1-2x+3\right)\left(3x-1+2x-3\right)\)

\(=\left(x+2\right)\left(5x-4\right)\)

d, \(5x-3x-2=2x-2=2\left(x-1\right)\)

\(\left(x+2\right)\left(9x^2-6x+4\right)\)

\(=9x^3-6x^2+4x+18x^2-12x+8\)

\(=9x^3+12x^2-8x+8\)

a, \(\left|x+1\right|-4=0\)

\(\Rightarrow\left|x+1\right|=4\)

+) Xét \(x\ge-1\) có:

\(x+1=4\Rightarrow x=3\) ( t/m )

+) Xét x < -1 có:

\(-x-1=4\Rightarrow x=-5\) ( t/m )

Vậy x = 3 hoặc x = -5

b, tương tự

c, \(3x+\left|2x\right|=5x\)

\(\Rightarrow\left|2x\right|=2x\)

+) Xét \(x\ge0\)

\(\Rightarrow2x=2x\Rightarrow x\in R\forall x>0\)

+) Xét x < 0

\(\Rightarrow-2x=2x\Rightarrow x=0\)

Vậy \(x\ge0\)

d, \(\left|2x+1\right|=\left|x-3\right|\)

\(\Rightarrow\left[{}\begin{matrix}2x+1=x-3\\2x+1=3-x\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-4\\x=\dfrac{2}{3}\end{matrix}\right.\)

Vậy...

Ta có: \(a+b+\dfrac{1}{2}\ge\sqrt{a}+\sqrt{b}\)

\(\Leftrightarrow\left(a-2\sqrt{a}.\dfrac{1}{2}+\dfrac{1}{4}\right)+\left(b-2\sqrt{b}.\dfrac{1}{2}+\dfrac{1}{4}\right)\ge0\)

\(\Leftrightarrow\left(\sqrt{a}-\dfrac{1}{2}\right)^2+\left(\sqrt{b}-\dfrac{1}{2}\right)^2\ge0\) ( luôn đúng )

\(\Rightarrowđpcm\)

\(A=-1^2+2^2-3^2+4^2-...-99^2+100^2\)

\(=\left(2-1\right)\left(2+1\right)+\left(4-3\right)\left(4+3\right)+...+\left(100-99\right)\left(100+99\right)\)

\(=3+7+...+199\)

\(=5050\)

\(A=3x^2+y^2-2x+y\)

\(=3\left(x^2-\dfrac{1}{3}x.2+\dfrac{1}{9}-\dfrac{1}{9}\right)+\left(y^2+2.y.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}\right)\)

\(=3\left(x-\dfrac{1}{3}\right)^2-\dfrac{1}{3}+\left(y+\dfrac{1}{2}\right)^2-\dfrac{1}{4}\)

\(=3\left(x-\dfrac{1}{3}\right)^2+\left(y+\dfrac{1}{2}\right)^2-\dfrac{7}{12}\ge\dfrac{-7}{12}\)

Dấu " = " khi \(\left\{{}\begin{matrix}3\left(x-\dfrac{1}{3}\right)^2=0\\\left(y+\dfrac{1}{2}\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{3}\\y=\dfrac{-1}{2}\end{matrix}\right.\)

Vậy \(MIN_A=\dfrac{-7}{12}\) khi \(\left\{{}\begin{matrix}x=\dfrac{1}{3}\\y=\dfrac{-1}{2}\end{matrix}\right.\)

Bài 1:

a, Ta có: \(3^3\equiv-1\left(mod28\right)\)

\(\Rightarrow3^{1179}\equiv-1\left(mod28\right)\)

\(\Rightarrow3^{1181}\equiv-9\left(mod28\right)\)

Vậy \(3^{1181}\) chia 28 dư -9

Bài 2:

\(2^5\equiv1\left(mod31\right)\)

\(\Rightarrow2^{2000}\equiv1\left(mod31\right)\)

\(\Rightarrow2^{2002}\equiv4\left(mod31\right)\)

\(\Rightarrow2^{2002}-4⋮31\)

Cách 2:

\(A=x^4-12x^3+12x^2-12x+111\)

\(=x^4-11x^3-x^3+11x^2+x^2-11x-x+11+100\)

\(=x^3\left(x-11\right)-x^2\left(x-11\right)+x\left(x-11\right)-\left(x-11\right)+100\)

\(=\left(x^3-x^2+x-1\right)\left(x-11\right)+100\)

Thay x = 11

\(\Rightarrow A=100\)

Vậy...

\(E=\dfrac{10^{15}+5}{10^{15}-8}=\dfrac{10^{15}-8+13}{10^{15}-8}=1+\dfrac{13}{10^{15}-8}\)

\(D=\dfrac{10^{16}+6}{10^{16}-7}=1+\dfrac{13}{10^{16}-7}\)

Vì \(\dfrac{13}{10^{15}-8}>\dfrac{13}{10^{16}-7}\Rightarrow1+\dfrac{13}{10^{15}-8}>1+\dfrac{13}{10^{16}-7}\)

\(\Rightarrow E>D\)

\(\Rightarrowđpcm\)

Bài 1:

Giải:

Đổi \(25\%=\dfrac{1}{4}\)

Phân số chỉ 42 kg gạo là:

\(1-\dfrac{1}{4}=\dfrac{3}{4}\) ( tổng số gạo )

Tổng số gạo là:
\(42:\dfrac{3}{4}=52\left(kg\right)\)

Số gạo bán ra lần đầu là:
\(52.\dfrac{1}{4}=13\left(kg\right)\)

Vậy...

Bài 2:

\(A=\dfrac{1}{1.4}+\dfrac{1}{4.7}+...+\dfrac{1}{2008.2011}\)

\(=\dfrac{1}{3}\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{2008.2011}\right)\)

\(=\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{2008}-\dfrac{1}{2011}\right)\)

\(=\dfrac{1}{3}\left(1-\dfrac{1}{2011}\right)\)

\(=\dfrac{1}{3}-\dfrac{1}{6033}< 1\)

Bài 3:

Đặt \(UCLN\left(12n+1;30n+2\right)=d\left(d\in Z\right)\)

\(\Rightarrow\left\{{}\begin{matrix}12n+1⋮d\\30n+2⋮d\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}60n+5⋮d\\60n+6⋮d\end{matrix}\right.\Rightarrow1⋮d\)

\(\Rightarrow d\in\left\{1;-1\right\}\)

\(\Rightarrow\left(12n+1;30n+2\right)=1\)

\(\Rightarrow\dfrac{12n+1}{30n+2}\) tối giản

Vậy...