Bài 5 : 

a, \(2x\left(x-3\right)+x-3=0\Leftrightarrow\left(2x+1\right)\left(x-3\right)=0\Leftrightarrow x=-\frac{1}{2};x=3\)

b, \(x\left(x+1\right)-x-1=0\Leftrightarrow\left(x-1\right)\left(x+1\right)=0\Leftrightarrow x=\pm1\)

c, sửa đề  \(x^3-3x^2+x-3=0\Leftrightarrow x^2\left(x-3\right)+x-3=0\)

\(\Leftrightarrow\left(x^2+1>0\right)\left(x-3\right)=0\Leftrightarrow x=3\)

d, \(3x^2\left(2x-1\right)+1-4x^2=0\Leftrightarrow3x^2\left(2x-1\right)+\left(1-2x\right)\left(1+2x\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(3x^2-2x-1\right)=0\Leftrightarrow\left(2x-1\right)\left(3x+1\right)\left(x-1\right)=0\Leftrightarrow x=1;x=-\frac{1}{3};x=\frac{1}{2}\)

e, \(x^3+2x-x^2-2=0\Leftrightarrow x\left(x^2+2\right)-\left(x^2+2\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+2>0\right)=0\Leftrightarrow x=1\)

x=1 nha

\(A=-x^2-x+1=-\left(x^2+x-1\right)=-\left(x^2+x+\frac{1}{4}-\frac{5}{4}\right)\)

\(=-\left(x+\frac{1}{2}\right)^2+\frac{5}{4}\le\frac{5}{4}\)

Dấu ''='' xảy ra khi x = -1/2 

Vậy GTLN A là 5/4 khi x = -1/2 

\(-x^2+3x+1=-\left(x^2+2.\frac{3}{2}x+\frac{9}{4}-\frac{13}{4}\right)=-\left(x+\frac{3}{2}\right)^2+\frac{13}{4}\le\frac{13}{4}\)

Dấu ''='' xảy ra khi x = -3/2 

Vậy GTLN biểu thức trên là 13/4 khi x = -3/2 

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

\(\Leftrightarrow x^2+6x+9-x^2+4=11\Leftrightarrow6x+2=0\Leftrightarrow x=-\frac{1}{3}\)

b, \(x^2-6x-7=0\Leftrightarrow\left(x+1\right)\left(x-7\right)=0\Leftrightarrow x=-1;x=7\)

c, \(\left(2x+1\right)^2-\left(3x-2\right)^2=0\Leftrightarrow\left(-x+3\right)\left(5x-1\right)=0\Leftrightarrow x=\frac{1}{5};x=3\)

x=3 nha

Bài 3 : 

\(D=40^2-28^2+32^2+80.32=40^2+2.40.32+32^2-28^2\)

\(=\left(40+32\right)^2-28^2=72^2-28^2=\left(72+28\right)\left(72-28\right)=46.100=4600\)

4600 nha

Bài 4 : 

\(B=\left(9+1\right)\left(9^2+1\right)...\left(9^{32}+1\right)\)

\(8B=\left(9^2-1\right)\left(9^2+1\right)...\left(9^{32}+1\right)=9^{64}-1\)

\(\Rightarrow B=\frac{9^{64}-1}{8}< 9^{64}-1\Rightarrow B< C\)

bài 5mn giúp em đc k ạ

a, \(A=x\left(3x+1\right)+3x+1=\left(x+1\right)\left(3x+1\right)\)

Thay x = 33 ta được : \(32.100=3200\)

b, \(B=xy+2x+2y+4=x\left(y+2\right)+2\left(y+2\right)=\left(x+2\right)\left(y+2\right)\)

Thay x = 98 ; y = 98 ta được : \(100.100=10000\)

10000 nha

Trả lời :

=> VD : \(\frac{2x+3}{3x^4-x^2+7}\)

\(\frac{2x+3}{3x^4-x^2-7}\)

=> Là phân thức đại số 

* Nguồn : Vietjack *