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17 tháng 2 2020

a/ (3x - 2)(4x + 5) = 0
<=> 3x - 2 = 0
       4x + 5 = 0
<=> 3x = 2
       4x = -5
<=> x = 2/3
       x = -4/5
Vậy pt có tập nghiệm là S = {2/3;-4/5}

a: A=(4x+5)^2-2*(4x+5)(4x-5)+(4x-5)^2

=(4x+5-4x+5)^2

=10^2=100

b:  B=(3x-2)^2*(3x+2)^2-2(2x+3)(2x-3)

=(9x^2-4)^2-2(4x^2-9)

=81x^4-72x^2+16-8x^2+18

=81x^4-80x^2+34

3 tháng 9 2023

\(a,A=\left(4x-5\right)^2+\left(4x+5\right)^2+2\left(5+4x\right)\left(5-4x\right)\)

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

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

\(=10^2\)

\(=100\)

\(b,B=\left(3x-2\right)^2\left(3x+2\right)^2-2\left(2x+3\right)\left(2x-3\right)\)

\(=\left(9x^2-4\right)^2-2\left(4x^2-9\right)\)

\(=81x^4-72x^2+16-8x^2+18\)

\(=81x^4-80x^2+34\)

#\(Urushi\)

30 tháng 8 2016

làm khuyến mại 1 câu;

a) = 12x2 -12x2 +20x -10x +17 =0

10x = -17

x = -17/10

5 tháng 9 2016

x/2 - ( 3x/5 - 13/5 ) = -( 7/5 + 7/10x )

14 tháng 9 2018

a ) \(\left(3x^2-4x+5\right)\left(2x^2-4\right)-2x\left(3x^3-4x^2+8\right)\)

\(=\left(3x^2-4x+5\right).2x^2-4\left(3x^2-4x+5\right)-6x^4+8x^3-16x\)

\(=6x^4-8x^3+10x^2-12x^2+16x-20-6x^4+8x^3-16x\)

\(=\left(6x^4-6x^4\right)+\left(8x^3-8x^3\right)-\left(12x^2-10x^2\right)+\left(16x-16x\right)-20\)

\(=-2x^2-20\)

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

\(=2\left(1-3x+x^2\right)-4x\left(1-3x+x^2\right)+4x^3+10x\)

\(=2-6x+2x^2-4x+12x^2-4x^3+4x^3+10x\)

\(=\left(4x^3-4x^3\right)+\left(12x^2+2x^2\right)+\left(10x-6x-4x\right)+2\)

\(=14x^2+2\)

31 tháng 5 2021

\(A=-x^2+3x-5\)\(=-\dfrac{11}{4}-\left(x^2-2.\dfrac{3}{2}x+\dfrac{9}{4}\right)=-\dfrac{11}{4}-\left(x-\dfrac{3}{2}\right)^2\le-\dfrac{11}{4}\) với mọi x

\(\Rightarrow A_{max}=-\dfrac{11}{4}\Leftrightarrow x-\dfrac{3}{2}=0\Leftrightarrow x=\dfrac{3}{2}\)

\(B=5x-4x^2-3=-\dfrac{23}{16}-\left(4x^2-2.\dfrac{5}{4}.2x+\dfrac{25}{16}\right)\)\(=-\dfrac{23}{16}-\left(2x-\dfrac{5}{4}\right)^2\)\(\le-\dfrac{23}{16}\forall x\)

\(\Rightarrow B_{max}=-\dfrac{23}{16}\Leftrightarrow2x-\dfrac{5}{4}=0\Leftrightarrow x=\dfrac{5}{8}\)

\(C=5-4x-25x^2=\dfrac{129}{25}-\left(25x^2+2.5x.\dfrac{2}{5}+\dfrac{4}{25}\right)\)\(=\dfrac{129}{25}-\left(5x+\dfrac{2}{5}\right)^2\le\dfrac{129}{25}\forall x\)

\(\Rightarrow C_{max}=\dfrac{129}{25}\Leftrightarrow5x+\dfrac{2}{5}=0\Leftrightarrow x=-\dfrac{2}{25}\)

31 tháng 5 2021

\(D=3x-2x^2=-2\left(x^2-\dfrac{3}{2}x\right)=-2\left(x^2-2.\dfrac{3}{4}x+\dfrac{9}{16}\right)+\dfrac{9}{8}\)\(=\dfrac{9}{8}-2\left(x-\dfrac{3}{4}\right)^2\le\dfrac{9}{8}\) với mọi x

\(\Rightarrow D_{max}=\dfrac{9}{8}\Leftrightarrow x-\dfrac{3}{4}=0\Leftrightarrow x=\dfrac{3}{4}\)

\(E=2+6x-\dfrac{1}{4}x^2=-\dfrac{1}{4}\left(x^2-24x\right)+2=-\dfrac{1}{4}\left(x^2-2.12x+144\right)+38\)\(=38-\dfrac{1}{4}\left(x-12\right)^2\le38\forall x\)

\(\Rightarrow E_{max}=38\Leftrightarrow x-12=0\Leftrightarrow x=12\)

\(F=-5x^2+4x=-5\left(x^2-\dfrac{4}{5}x\right)=-5\left(x^2-2.\dfrac{2}{5}x+\dfrac{4}{25}\right)+\dfrac{4}{5}\)\(=\dfrac{4}{5}-5\left(x-\dfrac{2}{5}\right)^2\le\dfrac{4}{5}\forall x\)

\(\Rightarrow F_{max}=\dfrac{4}{5}\Leftrightarrow x-\dfrac{2}{5}=0\Leftrightarrow x=\dfrac{2}{5}\)

5 tháng 7 2018

1) a) \(\left(3x-1\right)\left(9x^2+3x+1\right)-4x\left(x-5\right)\)

\(=27x^3+9x^2+3x-9x^2-3x-1-4x^2+20x\)

\(=27x^3+\left(9x^2-9x^2-4x^2\right)+\left(3x-3x+20x\right)+\left(-1\right)\)

\(=27x^3-4x^2+20x-1\)

b)\(\left(7x+2\right)\left(3-4x\right)-\left(x+3\right)\left(x^2-3x+9\right)\)

\(=21x-28x^2+6-8x-x^3+3x^2-9x-3x^2+9x-27\)

\(=\left(21x-8x-9x+9x\right)+\left(-28x^2+3x^2-3x^2\right)\)\(+\left(6-27\right)\)\(+\left(-x^3\right)\)

\(=13x-28x^2-21-x^3\)

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

\(=16x^2-12x+12x-9-8-4x-2x^2+4x+2x^2+x^3\)

\(=\left(16x^2-2x^2+2x^2\right)+\left(-12x+12x-4x+4x\right)\)\(+\left(-9-8\right)\)\(+x^3\)

\(=16x^2-17+x^3\)

d)\(\left(3x-8\right)\left(-5x+6\right)-\left(4x+1\right)\left(3x-2\right)\)

\(=-15x^2+18x+40x-48-12x^2+8x-3x+2\)

\(=\left(-15x^2-12x^2\right)+\left(18x+40x+8x-3x\right)\)\(+\left(-48+2\right)\)

\(=-27x^2+63x-46\)

e)\(\left(3x-6\right)4x-2x\left(3x+5\right)-4x^2\)

\(=12x^2-24x-6x^2-10x-4x^2\)

\(=\left(12x^2-6x^2-4x^2\right)+\left(-24x-10x\right)\)

\(=2x^2-34x\)

f)\(\left(5x-6\right)\left(6x-5\right)-x\left(3x+10\right)\)

\(=30x^2-25x-36x+30-3x^2-10x\)

\(=\left(30x^2-3x^2\right)+\left(-25x-36x-10x\right)+30\)

\(=27x^2-71x+30\)

5 tháng 7 2018

2) a)\(x\left(x+3\right)-x^2=6\)

\(\Rightarrow x^2+3x-x^2=6\)

\(\Rightarrow\left(x^2-x^2\right)+3x=6\)

\(\Rightarrow3x=6\)

\(\Rightarrow x=2\)

Vậy x=2

b) \(2x\left(x-5\right)+x\left(-2x-1\right)=6\)

\(\Rightarrow2x^2-10x-2x^2-x=6\)

\(\Rightarrow\left(2x^2-2x^2\right)+\left(-10x-x\right)=6\)

\(\Rightarrow-11x=6\)

\(\Rightarrow x=-\dfrac{6}{11}\)

\(\)Vậy \(x=-\dfrac{6}{11}\)

c) x(x+5)-(x+1)(x-2)=7

\(\Rightarrow x^2+5x-x^2+2x-x+2=7\)

\(\Rightarrow\left(x^2-x^2\right)+\left(5x+2x-x\right)=7-2\)

\(\Rightarrow6x=5\)

\(\Rightarrow x=\dfrac{5}{6}\)

Vậy x=\(\dfrac{5}{6}\)

d)\(\left(3x+4\right)\left(6x-3\right)-\left(2x+1\right)\left(9x-2\right)=10\)

\(\Rightarrow18x^2-9x+24x-12-18x^2+4x-9x+2=10\)

\(\Rightarrow\left(18x^2-18x^2\right)+\left(-9x+24x+4x-9x\right)+\left(-12+2\right)=10\)

\(\Rightarrow10x-10=10\)

\(\Rightarrow10x=20\)

\(\Rightarrow x=2\)

Vậy x=2

a: \(\Leftrightarrow9x^2-12x+4-6x^2-16x=0\)

\(\Leftrightarrow3x^2-28x+4=0\)

\(\text{Δ}=\left(-28\right)^2-4\cdot3\cdot4=736>0\)

Do đó: Phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{28-4\sqrt{46}}{6}=\dfrac{14-2\sqrt{46}}{3}\\x_2=\dfrac{14+2\sqrt{46}}{3}\end{matrix}\right.\)

b: \(\Leftrightarrow16x^2+24x+9-16x^2+25=12\)

=>24x+34=12

=>24x=-22

hay x=-11/12