a) (x-1):2/3=-2/5

=>x-1=-4/15

=>x=11/15

b) |x-1/2|-1/3=0

=>|x-1/2|=1/3

=>\(\left\{{}\begin{matrix}x=\dfrac{1}{3}+\dfrac{1}{2}=\dfrac{5}{6}\\x=-\dfrac{1}{3}+\dfrac{1}{2}=\dfrac{1}{6}\end{matrix}\right.\) 

c) Tương Tự câu B

a: Ta có: \(\left(5x+1\right)^2-\left(5x-3\right)\left(5x+3\right)=30\)

\(\Leftrightarrow25x^2+10x+1-25x^2+9=30\)

\(\Leftrightarrow10x=20\)

hay x=2

b: Ta có: \(\left(x-1\right)\left(x^2+x+1\right)-x\left(x+2\right)\left(x-2\right)=5\)

\(\Leftrightarrow x^3-1-x^3+4x=5\)

\(\Leftrightarrow4x=6\)

hay \(x=\dfrac{3}{2}\)

a)\(3x^2-4x=0<=>x(3x-4)=0\)
TH1: x=0

TH2 3x-4=0 <=>x=4/3

KL:.....

b) .

<=> (x+3)(x-1+2x)=0

TH1: x+3=0 <=> x=-3

TH2  x-1=0  <=> x=1

KL:.....

c) \(9x^2+6x+1=0. <=>(3x+1)^2=0<=>3x+1=0<=>x=-1/3 ​\)

KL:......
d) \(x^2−4x=4.<=>(x-2)^2=0<=>x-2=0<=>x=2\)

KL:....

a) \(3x^2-4x=0\)

\(\Leftrightarrow x\left(3x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4}{3}\end{matrix}\right.\)

b) \(\left(x+3\right)\left(x-1\right)+2x\left(x+3\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{1}{3}\end{matrix}\right.\)

c) \(9x^2+6x+1=0\)

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

\(\Leftrightarrow3x+1=0\Leftrightarrow x=-\dfrac{1}{3}\)

d) \(x^2-4x=4\)

\(\Leftrightarrow\left(x-2\right)^2=8\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=2\sqrt{2}\\x-2=-2\sqrt{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\sqrt{2}+2\\x=-2\sqrt{2}+2\end{matrix}\right.\)

\(\left|x-2016\right|+\left|1008-\frac{1}{2}y\right|=0\)

\(\Leftrightarrow\begin{cases}x-2016=0\\1008-\frac{1}{2}y=0\end{cases}\)\(\Leftrightarrow x=y=2016\)

\(\left|x-2016\right|+\left|1008-\frac{1}{2}y\right|=0\)

\(\Rightarrow\left|x-2016\right|=0\) và \(\left|1008-\frac{1}{2}y\right|=0\)

+) \(\left|x-2016\right|=0\Rightarrow x-2016=0\Rightarrow x=2016\)

+) \(\left|1008-\frac{1}{2}y\right|=0\)

\(\Rightarrow1008-\frac{1}{2}y=0\)

\(\Rightarrow\frac{1}{2}y=1008\)

\(\Rightarrow y=2016\)

Vậy \(x=y=2016\)

1a) (x - 2y) (x² - 2xy + y²)

= (x - 2y) (x - y)²

= x² - xy - 2xy + 2y²

= (x² - xy) - (2xy - 2y²)

= x (x - y) - 2y (x - y)

= (x - y) (x - 2y)

2a) x (x - 3) - y (3 - x)

= x (x - 3) + y (x - 3)

= (x - 3) (x + y)

b) 3x² - 5x - 3xy + 5y

= (3x² - 3xy) - (5x - 5y)

= 3x (x - y) - 5 (x - y)

= (x - y) (3x - 5)

3) 12x (3 - 4x) + 7 (4x - 3) = 0

12x (3 - 4x) - 7 (3 - 4x) = 0

(3 - 4x) (12x - 7) = 0

=> 3 - 4x = 0 hoặc 12x - 7 = 0

* 3 - 4x = 0 => x = \(\frac{3}{4}\)

* 12x - 7 = 0 => x = \(\frac{7}{12}\)

Vậy x =\(\frac{3}{4}\)hoặc x =\(\frac{7}{12}\)

\(A=x^6+2x\left(x^2+y\right)+x^2+y^2+26\) 

   \(=x^6+2x^2+2xy+x^2+y^2+26\) 

    \(=x^6+2x^2+\left(x+y\right)^2+26\ge26\forall x;y\) 

Dấu "=" xảy ra<=> \(x=0\) và \(\left(x+y\right)^2=0\Rightarrow y=0\) 

Vậy A_min =26 tại x=y=0

B=\(y^2-2xy+3x^2+2y-14x+1949\)

 \(=\left(y^2-2xy+x^2+2y-2x+1\right)+\left(2x^2-12x+18\right)+1930\)

 \(=\left(x-y-1\right)^2+2\left(x-3\right)^2+1930\)

  \(\ge1930\)

MinB=1930 khi \(\hept{\begin{cases}x=y+1\\x=3\end{cases}\Rightarrow\hept{\begin{cases}x=3\\y=2\end{cases}}}\)

D=\(x\left(x-1\right)\left(x-2\right)\left(x-3\right)\)

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

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

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

MinD=-1 khi \(x^2-3x+1=0\)=> \(x=\frac{3\pm\sqrt{5}}{2}\)

b)\(\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}=3\left(x+y\right)\)

\(\Rightarrow\left(\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}\right)^2=\left(3\left(x+y\right)\right)^2\)

\(\Leftrightarrow\sqrt{\left(5x^2+2xy+2y^2\right)\left(2x^2+2xy+5y^2\right)}=x^2+7xy+y^2\)

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

\(\Leftrightarrow9\left(x-y\right)^2\left(x+y\right)^2=0\)\(\Leftrightarrow\left[{}\begin{matrix}x=y\\x=-y\end{matrix}\right.\)

\(\rightarrow\left(x;y\right)\in\left\{\left(0;0\right),\left(1;1\right)\right\}\)

caau a) binh phuong len ra no x=y tuong tu

c)

ĐK $y \geqslant 0$

Hệ đã cho tương đương với

$\left\{\begin{matrix} 2x^2+2xy+2x+6=0\\ (x+1)^2+3(y+1)+2xy=2\sqrt{y(x^2+2)} \end{matrix}\right.$

Trừ từng vế $2$ phương trình ta được

$x^2+2+2\sqrt{y(x^2+2)}-3y=0$

$\Leftrightarrow (\sqrt{x^2+2}-\sqrt{y})(\sqrt{x^2+2}+3\sqrt{y})=0$

$\Leftrightarrow x^2+2=y$