a)để  \(2x^2-4x\)dương

\(\Leftrightarrow2x^2-4x>0\)

\(\Leftrightarrow2x\left(x-2\right)>0\)

TH1: \(\Rightarrow\hept{\begin{cases}2x>0\\x-2>0\end{cases}\Rightarrow\orbr{\begin{cases}x>0\\x>2\end{cases}}\Rightarrow x>2}\)

TH2: \(\hept{\begin{cases}2x< 0\\x-2< 0\end{cases}\Rightarrow\orbr{\begin{cases}x< 0\\x< 2\end{cases}}\Rightarrow x< 0}\)

ta có: 2(x-3) - 3(1-2x)=4+4(1-x)

=>    2x-6 - 3+6x = 4+4 - 4x

=>    2x+6x+4x= 4 + 4 +6+3

=>    12x         = 17

          x= 17/12

chúc bạn học tốt nhé!

\(3\left(2x-6\right)-4\left(1+2x\right)-2\left(x-4\right)=4-3\left(1+2x\right)-5\left(1-2x\right).\)

\(\Leftrightarrow6x-18-4-8x-2x+8=4-3-6x-5+10x\)

\(\Leftrightarrow-4x-14=4x-4\)

\(\Leftrightarrow-4x-4x=-4+14\)

\(\Leftrightarrow-8x=10\)

\(\Leftrightarrow x=-\frac{5}{4}\)

a/ \(=\sqrt{\left(\sqrt{3}-1\right)^2\left(2\sqrt{3}+1\right)^2}\)

\(=\left(\sqrt{3}-1\right)\left(2\sqrt{3}+1\right)=5-\sqrt{3}\)

b/ \(=\left(\sqrt{3}-2\right)\left(\sqrt{3}+1\right)\sqrt{4+2\sqrt{3}}\)

\(=\left(\sqrt{3}-2\right)\left(\sqrt{3}+1\right)\sqrt{\left(\sqrt{3}+1\right)^2}\)

\(=\left(\sqrt{3}-2\right)\left(\sqrt{3}+1\right)\left(\sqrt{3}+1\right)=\left(\sqrt{3}-2\right)\left(\sqrt{3}+1\right)^2\)

\(=\left(\sqrt{3}-2\right)\left(4+2\sqrt{3}\right)=2\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)\)

\(=2\left(3-4\right)=-2\)

c/ \(=\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\sqrt{6-2\sqrt{5}}\)

\(=\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\sqrt{\left(\sqrt{5}-1\right)^2}\)

\(=\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)^2=\left(3+\sqrt{5}\right)\left(6-2\sqrt{5}\right)\)

\(=2\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)=2.\left(9-5\right)=8\)

d/ \(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{8-2\sqrt{15}}\)

\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)

\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)^2=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)

\(=2\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)=2\left(16-15\right)=2\)

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

\(\Leftrightarrow4x^2-4x+1-3\left(x^2+4x+4\right)=4x-8-5.\left(x^2-2x+1\right)\)

\(\Leftrightarrow4x^2-4x+1-3x^2-7x-12=4x-8-5x^2+10x-5\)

\(\Leftrightarrow x^2-11x-11=14x-13-5x^2\)

\(\Leftrightarrow6x^2-25x+2=0\)

Tự làm tiếp nha

~~~~~~~~~~~ai đi ngang qua nhớ để lại k ~~~~~~~~~~~~~

 ~~~~~~~~~~~~ Chúc bạn sớm kiếm được nhiều điểm hỏi đáp ~~~~~~~~~~~~~~~~~~~

~~~~~~~~~~~ Và chúc các bạn trả lời câu hỏi này kiếm được nhiều k hơn ~~~~~~~~~~~~

bạn giải tiếp giúp mk với được ko

Giải tới đây pt có 2 ngiệm\(\hept{\begin{cases}x_1=\frac{25+\sqrt{577}}{12}\\x_2=\frac{25-\sqrt{577}}{12}\end{cases}}\)

b)(2x - 1)^2 - (2x + 5) (2x - 5 ) = 18

4x 2 -4x+1-4x 2+25=18

26-4x=18

4x=8

x=2

a,27x-18=2x-3x^2

<=> 3x^2-2x+27-18x=0

<=> 3x^2-20x+27=0

\(\Delta\)= 20^2-4-12.27

tính \(\Delta\)rồi tìm x1 ,x2

â)\(9\left(3x-2\right)=x\left(2-3x\right)\)

\(\Leftrightarrow27x-18=2x-3x^2\)

\(\Leftrightarrow27x-18-2x+3x^2=0\)

\(\Leftrightarrow3x^2+25x-18=0\)

\(\Leftrightarrow3x^2+27x-2x-18=0\)

\(\Leftrightarrow\left(3x-2\right)\left(x+9\right)=0\)

       \(\Rightarrow\orbr{\begin{cases}3x-2=0\\x+9=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{2}{3}\\x=-9\end{cases}}\)

b)\(\left(2x-1\right)^2-\left(2x+5\right)\left(2x-5\right)=18\)

\(\Leftrightarrow4x^2-4x+1-4x^2+25=18\)

\(\Leftrightarrow26-4x=18\)

 \(\Leftrightarrow4x=8\)

      \(\Rightarrow x=2\)

c)\(5x\left(x-5\right)-2x+10=0\)

\(\Leftrightarrow5x^2-10x-2x+10=0\)

\(\Leftrightarrow5x^2-12x+10=0\)

\(\Leftrightarrow x^2-6x+2=0\)

\(\Leftrightarrow x^2-6x+9-7=0\)

\(\Leftrightarrow\left(x-3\right)^2=7\)

       \(\Rightarrow\orbr{\begin{cases}x-3=\sqrt{7}\\x-3=-\sqrt{7}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\sqrt{7}+3\\x=-\sqrt{7}+3\end{cases}}\)

d)\(x^2-5=0\)

\(\Leftrightarrow x^2=5\)

    \(\Rightarrow x=\sqrt{5};-\sqrt{5}\)

e)\(x^3+5x^2-4x-20=0\)

\(\Leftrightarrow x^3-2x^2+7x^2-14x+10x-20=0\)

\(\Leftrightarrow x^2\left(x-2\right)+7x\left(x-2\right)+10\left(x-2\right)=0\)

\(\Leftrightarrow\left(x^2+7x+10\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(x+5\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left(x^2-4\right)\left(x+5\right)=0\)

      \(\Rightarrow\orbr{\begin{cases}x+5=0\\x^2-4=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=-5\\x^2=4\end{cases}\Rightarrow}\orbr{\begin{cases}x=-5\\x=-2;2\end{cases}}\)

1/ đkxđ: x≠\(\pm\)1; x≠1/2

a/\(A=\left(\dfrac{1}{1-x}+\dfrac{2}{x+1}-\dfrac{5-x}{1-x^2}\right):\dfrac{1-2x}{x^2-1}\)

\(=\left(\dfrac{x+1}{\left(1-x\right)\left(1+x\right)}+\dfrac{2\left(1-x\right)}{\left(1-x\right)\left(1+x\right)}-\dfrac{5-x}{\left(1-x\right)\left(1+x\right)}\right)\cdot\dfrac{\left(x-1\right)\left(x+1\right)}{1-2x}\)

\(=\dfrac{x+1+2-2x-5+x}{\left(1-x\right)\left(1+x\right)}\cdot\dfrac{\left(x-1\right)\left(x+1\right)}{1-2x}\)

\(=\dfrac{2}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{\left(x-1\right)\left(x+1\right)}{1-2x}=\dfrac{2}{1-2x}\)

b/ A nguyên <=> 1 - 2x ∈ Ư(2)

<=> 1 - 2x = {-2;-1;1;2}

<=> -2x = {-3; -2; 0;1}

<=> x = {3/2; 1; 0; -1/2}

mà x nguyên => x = {1;0}

c/ \(\left|A\right|=A\Leftrightarrow\left|\dfrac{2}{1-2x}\right|=\dfrac{2}{1-2x}\)

+) Với x > 1/2 có:

\(\dfrac{2}{1-2x}=\dfrac{2}{1-2x}\Leftrightarrow\dfrac{2}{1-2x}-\dfrac{2}{1-2x}=0\Leftrightarrow0x=0\)

=> x>1/2 thỏa mãn là nghiệm

+) Với x < 1/2 có:

\(\dfrac{2}{1-2x}=\dfrac{2}{2x-1}\)

\(\Leftrightarrow\dfrac{2}{1-2x}-\dfrac{2}{2x-1}=0\Leftrightarrow\dfrac{2}{1-2x}+\dfrac{2}{1-2x}=0\)

\(\Leftrightarrow\dfrac{4}{1-2x}=0\) mà 1 - 2x ≠ 0 => vô nghiệm

Vậy x>1/2

Ta có : A = x(x + 1)(x² +  x - 4)

= (x² + x)(x² + x - 4)

Đặt x² + x = t

Khi đó A = t(t - 4)

= t² - 4t = t² - 4t + 4 - 4 = (t - 2)² - 4 \(\ge\)-4

 Dấu "=" xảy ra <=> t - 2 = 0

=> t = 2

=> x² + x = 2

=> x² + x - 2 = 0

=> x² + 2x - x - 2 = 0

=> x(x + 2) - (x + 2) = 0

=> (x - 1)(x + 2) = 0

=> \(\orbr{\begin{cases}x-1=0\\x+2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=-2\end{cases}}\)

Vậy Min A = -4 <=> x \(\in\left\{1;-2\right\}\)

A = x( x + 1 )( x² + x - 4 )

= ( x² + x )( x² + x - 4 )

Đặt t = x² + x

A <=> t( t - 4 )

      = t² - 4t

      = ( t² - 4t + 4 ) - 4

      = ( t - 2 )² - 4 

      = ( x² + x - 2 )² - 4 ≥ -4 ∀ x

Đẳng thức xảy ra <=> x² + x - 2 = 0

                             <=> x² - x + 2x - 2 = 0

                             <=> x( x - 1 ) + 2( x - 1 ) = 0

                             <=> ( x - 1 )( x + 2 ) = 0

                             <=> \(\orbr{\begin{cases}x-1=0\\x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-2\end{cases}}\)

=> MinA = -4 <=> x = 1 hoặc x = -2

a,\(A=x\left(x+1\right)\left(x^2+x-4\right)\)

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

Đặt \(x^2+x=t\)ta có:

\(A=t\left(t-4\right)\)

\(=t^2-4t\)

\(=\left(t^2-4t+4\right)-4\)

\(=\left(t-2\right)^2-4\ge-4\forall t\)

Dấu "="xảy ra khi \(\left(t-2\right)^2=0\Rightarrow t=2\)

\(\Rightarrow Min_A=-4\Leftrightarrow t=2\)

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

b,\(B=-x^2-y^2+xy+2x+2y\)

\(\Leftrightarrow-2B=2x^2+2y^2-2xy-4x-4y\)

\(=\left(x^2-2xy+y^2\right)+\left(x^2-4x+4\right)+\left(y^2-4y+4\right)-8\)

\(=\left(x-y\right)^2+\left(x-2\right)^2+\left(y-2\right)^2-8\ge-8\Leftrightarrow B\le4\)

Dấu"="xảy ra khi \(\hept{\begin{cases}\left(x-y\right)^2=0\\\left(x-2\right)^2=0\\\left(y-2\right)^2=0\end{cases}\Rightarrow x=y=2}\)

Vậy \(Max_B=4\Leftrightarrow x=y=2\)