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\(\sqrt{x}+2\sqrt{y}=10=>\left(\sqrt{x}+2\sqrt{y}\right)^2=100\)

BDT Bunhiacopxki (đề sai phải lớn hơn bằng 20)

\(=>\left(\sqrt{x}+2\sqrt{y}\right)^2\le\left(1^2+2^2\right)\left(x+y\right)=5\left(x+y\right)\)

\(< =>5\left(x+y\right)\ge100=>x+y\ge20\)

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

a) x + 10 = 20

<=> x = 20 - 10 = 10

Vậy x = 10

b) 2x + 15 = 35

<=> 2x = 35 - 15 = 20

<=> x = 10

Vậy x = 10

c) 3(x + 2) = 15

<=> x + 2 = 15 : 3 = 5

<=> x = 5 - 2 = 3

Vậy x = 3

d) 10x + 15.11 = 20.10

<=> 10x + 165 = 200

<=> 10x = 200 - 165 = 35

<=> x = 35 : 10 = 3,5

Vậy x = 3,5

e) 4(x + 2) = 3.4

<=> x + 2 = 3

<=> x = 3 - 2 = 1

Vậy x = 1

f) 33x + 135 = 26.9

<=> 33x + 135 = 234

<=> 33x = 234 - 135 = 99

<=> x = 99 : 33 = 3

Vậy x = 3

g) 2x + 15 + 16 + 17 = 100

<=> 2x + 48 = 100

<=> 2x = 100 - 48 = 52

<=> x = 52 : 2 = 26

Vậy x = 26

h) 2(x + 9 + 10 + 11) = 4.12.5

<=> x + 30 = 120

<=> x = 120 - 30 = 90

Vậy x = 90