k cần vẽ hình mk cũng cm dc

pg A cắt B tai k trg tg akb có k =90=> a+b =90

vay tg ABCD có  A+ B= 180( 2 góc này ở vi trí trong cùng phía )

nên AD//BC => abcd là hthang

\(\left(x-1\right)\left(x^2+x+1\right)-x\left(x^2-2\right)-7=x^3-1-x^3+2x-7\) (hằng đẳng thức \(x^3-y^3\) )

\(=-1+2x-7=-8+2x\)

Xem lại đề

\(Q=2x^2-6x=2\left(x^2-3x\right)=2\left(x^2-2.x.\frac{3}{2}+\frac{9}{4}-\frac{9}{4}\right)\)

\(=2\left[\left(x-\frac{3}{2}\right)^2-\frac{9}{4}\right]=2\left(x-\frac{3}{2}\right)^2-\frac{9}{2}\ge-\frac{9}{2}\)

Dấu "=" xảy ra \(< =>\left(x-\frac{3}{2}\right)^2=0< =>x=\frac{3}{2}\)

Vậy MInQ=-9/2 khi x=3/2

\(M=x^2+y^2-x+6y+10=x^2+y^2-x+6y+1+9=\left(x^2-x+1\right)+\left(y^2+6y+9\right)\)

\(=\left[\left(x^2-2.x.\frac{1}{2}+\frac{1}{4}\right)+\frac{3}{4}\right]+\left(y^2+2.y.3+9\right)=\left[\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\right]+\left(y+3\right)^2=\left(x-\frac{1}{2}\right)^2+\left(y+3\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

Dấu "=" xảy ra \(< =>\left(x-\frac{1}{2}\right)^2=0=>x=\frac{1}{2}\)

  và \(\left(y+3\right)^2=0=>y=-3\)

Vậy minM=3/4 khi x=1/2 và y=-3

Tìm giá trị nhỏ nhất nhé

Đặt \(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

\(=4\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

Ta có: \(2.A=8\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

\(=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

\(=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

\(=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)=\left(3^{16}-1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)=\left(3^{32}-1\right)\left(3^{32}+1\right)\)

\(=3^{64}-1=>A=\frac{3^{64}-1}{2}\)