thế còn đỡ mị bị viết bản tường trình chơi ĐT trong lp ms cay

\(x^4+x+1\)

\(=\left(x^4-x^3+x^2\right)+\left(x^3+1\right)\)

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

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

\(x^5+x+1\)

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

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

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

\(a\left(b^2-c^2\right)+b\left(c^2-a^2\right)+c\left(a^2-b^2\right)\)

\(=a\left(b-c\right)\left(b+c\right)+bc^2-a^2b+a^2c-b^2c\)

\(=a\left(b-c\right)\left(b+c\right)+\left(bc^2-b^2c\right)-\left(a^2b-a^2c\right)\)

\(=a\left(b-c\right)\left(b+c\right)+bc\left(c-b\right)-a^2\left(b-c\right)\)

\(=\left(b-c\right)\left(ab+ac-bc-a^2\right)\)

\(=\left(b-c\right)\left[\left(ac-a^2\right)+\left(ab-bc\right)\right]\)

\(=\left(b-c\right)\left[a\left(c-a\right)+b\left(a-c\right)\right]=\left(b-c\right)\left(c-a\right)\left(a-b\right)\)

Có: \(\frac{k}{x}=\frac{a}{c}\Rightarrow ax=kc\)

      \(\frac{k}{y}=\frac{b}{d}\Rightarrow by=kd\)

=> \(ax+by=kc+kd=k\left(c+d\right)=k\cdot k=k^2\)

=>đpcm

\(2\cdot2^2\cdot2^3\cdot2^4\cdot\cdot\cdot2^x=32768\)

\(\Leftrightarrow2^{1+2+3+4+\cdot\cdot\cdot+x}=2^{15}\)

\(\Leftrightarrow1+2+3+4+..+x=15\)

\(\Leftrightarrow\)\(\frac{\left(1+x\right)x}{2}=15\)

\(\Leftrightarrow x\left(x+1\right)=30=5\left(5+1\right)\)

Vậy x=5

Bài 2:

Bậc của đơn thức là 2+5+3=10

Bài 3:

\(\left|2x-\frac{1}{2}\right|+\frac{3}{7}=\frac{38}{7}\)

\(\Leftrightarrow\left|2x-\frac{1}{2}\right|=5\)

+)TH1: \(x\ge\frac{1}{4}\) thì bt trở thành

\(2x-\frac{1}{2}=5\Leftrightarrow2x=\frac{11}{2}\Leftrightarrow x=\frac{11}{4}\left(tm\right)\)

+)TH2: \(x< \frac{1}{4}\) thì pt trở thành

\(2x-\frac{1}{2}=-5\Leftrightarrow2x=-\frac{9}{2}\Leftrightarrow x=-\frac{9}{4}\left(tm\right)\)

Vậy x={-9/4;11/4}

Bài 1:

\(M=\left|x+13\right|+64\)

Vì \(\left|x+3\right|\ge0\)

=> \(\left|x+3\right|+64\ge64\)

Vậy GTNN của M là 64 khi x=-13

\(A=\left|x+3\right|+\left|x+5\right|=\left|-\left(x+3\right)\right|+\left|x+5\right|\)

Áp dụng bđt \(\left|A\right|+\left|B\right|\ge\left|A+B\right|\) ta có:

\(A\ge\left|-x-3+x+5\right|=2\)

Vaayj GTNN của A là 2 khi \(-3\le x\le5\)

Bài 2:

a) \(\left(x+10\right)^2=0\)

\(\Leftrightarrow x+10=0\Leftrightarrow x=-10\)

b) \(\left(x-\sqrt{121}\right)\left(x^2+1\right)=0\)

\(\Leftrightarrow x-\sqrt{121}=0\) (vì \(x^2+1>0\) )

\(\Leftrightarrow x=11\)

Bài 1:

\(\left(x+10\right)^2=0\)

\(\Leftrightarrow x+10=0\Leftrightarrow x=-10\)

Bài 2:

\(\left|x+13\right|+64\)

Vì \(\left|x+3\right|\ge0\)

=> \(\left|x+3\right|+64\ge64\)

Vậy GTNN của bt trên là 64 khi x=-3

\(A=\sqrt{x^2+2x+1}+\sqrt{x^2-2x+1}=\sqrt{\left(x+1\right)^2}+\sqrt{\left(x-1\right)^2}\)

\(=\left|x+1\right|+\left|x-1\right|=\left|x+1\right|+\left|1-x\right|\)

Áp dụng bđt \(\left|A\right|+\left|B\right|\ge\left|A+B\right|\) ta có:

\(A\ge\left|x+1+1-x\right|=2\)

Vậy GTNN của A là 2 khi \(-1\le x\le1\)

tốn S