Ta có: \(VT=\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+21}\)

\(=\sqrt{3x^2+6x+3+4}+\sqrt{5x^2+10x+5+16}\)

\(=\sqrt{3\left(x+1\right)^2+4}+\sqrt{5\left(x+1\right)^2+16}\ge2+4=6\)

Ta có: \(VP=5-x^2-2x\)

\(=-\left(x^2+2x+1\right)+6\)

\(=-\left(x+1\right)^2+6\le6\)

VP=VT khi x+1=0

hay x=-1

Vậy: x=-1

\((x-2)^3-(x+5)(x^2-5x+25)+6x^2=11\\\Leftrightarrow (x-2)^3-(x+5)(x^2-5.x+5^2)+6x^2=11 \\\Leftrightarrow x^3-6x^2+12x-8 -(x^3+5^3)+6x^2-11=0 \\\Leftrightarrow 12x-144=0 \\\Leftrightarrow x=12\)

Vậy \(x=12\).

(x−2)3−(x+5)(x2−5x+25)+6x2=11

=>(x−2)3−(x+5)(x2−5.x+52)+6x2=11

=>x3−6x2+12x−8−(x3+53)+6x2−11=0

=>12x−144=0

=>x=12(x−2)3−(x+5)(x2−5x+25)+6x2=11

=>(x−2)3−(x+5)(x2−5.x+52)+6x2=11

=>x3−6x2+12x−8−(x3+53)+6x2−11=0

=>12x−144=0

=>x=12

Vậy x=12x=12.

cho tôi đúng đi

b. 6x(x - 5) - x(6x + 3)

= x(6x - 30) - x(6x + 3)

= x(6x - 30 - 6x - 3)

= x(-33)

= -33x

\(1,\\ a,=-35x^5y^4z\\ b,=6x^2-30x-6x^2-3x=-33x\\ c,=x^3-9x^2-2x^2+18x-x+9=x^3-11x^2+17x+9\\ 2,\\ A\left(x\right)+B\left(x\right)=10-2x+4x^3-5x^2-10x^3-5x+6x^2-20\\ =-6x^3+x^2-7x-10\\ A\left(x\right)-B\left(x\right)=10-2x+4x^3-5x^2+10x^3+5x-6x^2+20\\ =14x^3-11x^2+3x+30\\ 3,\\ a,M\left(x\right)=5x+20=0\\ \Leftrightarrow x=-4\\ b,N\left(x\right)=100x^2-49=0\\ \Leftrightarrow\left(10x-7\right)\left(10x+7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{10}\\x=-\dfrac{7}{10}\end{matrix}\right.\\ c,P\left(x\right)=3x-15=0\\ \Leftrightarrow x=5\)

Bài 1;

a)\(5x^3yz.\left(-7x^2y^3\right)=-35.x^5y^4z\)

b)\(6x\left(x-5\right)-x\left(6x+3\right)=6x^2-30x-6x^2-3x=-33x\)

c) \(\left(x-9\right)\left(x^2-2x-1\right)=x^3-2x^2-x-9x^2+18x+9=x^3-11x^2+17x+9\)

                                                                     Bài giải

a, \(-\left(-x\right)-\left(-9\right)=-\left(3-6-9+3\right)\)

\(x+9=-3+6+9-3\)

\(x+9=15\)

\(x=15-9\)

\(x=6\)

b, \(17-x=7-6x\)

\(6x-x=7-17\)

\(5x=-10\)

\(x=-10\text{ : }5\)

\(x=-2\)

c, \(5x-7=-21-2x\)

\(5x+2x=-21+7\)

\(7x=-14\)

\(x=-14\text{ : }7\)

\(x=-2\)

d, \(7\left(x-3\right)-5\left(3-x\right)=11x-5\)

\(7x-21-15+5x=11x-5\)

\(7x+5x-11x=21+15-5\)

\(x=31\)

a) \(5x^3-125=0\)

\(\Leftrightarrow5x^3=125\)

\(\Leftrightarrow x^3=25\)

\(\Leftrightarrow x^3=25\)

\(\Leftrightarrow x=\sqrt[3]{25}\)

c) \(6x+13x+5=0\)

\(\Leftrightarrow19x+5=0\)

\(\Leftrightarrow19x=-5\)

\(\Leftrightarrow x=\dfrac{-5}{19}\)

a, \(5x^3-125=0\)

\(\Leftrightarrow5x^3=125\)

\(\Leftrightarrow x^3=25\)

\(\Leftrightarrow x=\sqrt[3]{25}\)

c, \(6x+13x+5=0\)

\(\Leftrightarrow19x+5=0\)

\(\Leftrightarrow19x=-5\)

\(\Leftrightarrow x=\dfrac{-5}{19}\)

a. (5x-1)²  -  (5x-4) (5x-4) +7

= (5x-1)² - (5x-4)²  + 7

=[(5x-1)+(5x-4)] [(5x-1)-(5x-4)] +7  ( đoạn này bỏ cx đc)

=(10x-5) .3+7

=30x-15+7

=30x-8

ý a hơi sai sai