\(\left(5x-3\right)^2-2\left(5x-3\right)=0\)

\(\left(5x-3\right)\left(5x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\\x=1\end{matrix}\right.\)

`(5x - 3)^2 - 10x - 6 = 0`

`<=> (5x - 3)(5x - 3) - 10x - 6 = 0`

`<=> 5x(5x - 3) - 3(5x - 3) - 10x - 6 = 0`

`<=> 25x^2 - 15x - 3(5x - 3) - 10x - 6 = 0`

`<=> x = (4 +- \sqrt{13})/(5)`

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

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

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

\(\Leftrightarrow\)\(\orbr{\begin{cases}x-5=0\\x-4=0\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=5\\x=4\end{cases}}\)

Vậy....

\(2x\left(x+6\right)=7x+42\)

\(\Leftrightarrow\)\(2x\left(x+6\right)-7x-42=0\)

\(\Leftrightarrow\)\(2x\left(x+6\right)-7\left(x+6\right)=0\)

\(\Leftrightarrow\)\(\left(x+6\right)\left(2x-7\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x+6=0\\2x-7=0\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-6\\x=\frac{7}{2}\end{cases}}\)

Vậy......

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

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

\(\Leftrightarrow\)\(\left(x-5\right)\left(x^2+1\right)=0\)

\(\Leftrightarrow\)\(x-5=0\)

\(\Leftrightarrow\)\(x=5\)

\(x^4-2x^3+10x^2-20x=0\)

\(\Leftrightarrow\)\(x^3\left(x-2\right)+10x\left(x-2\right)=0\)

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

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

Vậy...

\(\Leftrightarrow5\left(x^4+2x^2+1\right)+2\left(y^6+2y^3+1\right)=13\)

\(\Leftrightarrow5\left(x^2+1\right)^2+2\left(y^3+1\right)^2=13\)

\(\Leftrightarrow\left(x^2+1\right)^2=\dfrac{13-2\left(y^3+1\right)^2}{5}\le\dfrac{13}{5}< 4\)

\(\Rightarrow x^2+1< 2\Rightarrow x^2< 1\)

\(\Leftrightarrow x=0\)

\(\Rightarrow y^6+2y^3-3=0\Rightarrow\left[{}\begin{matrix}y^3=1\Rightarrow y=1\\y^3=-3\left(ktm\right)\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(0;1\right)\)

1.

<=> \(\left[{}\begin{matrix}4-3x=0\\10-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=2\end{matrix}\right.\)

2.

<=>\(\left[{}\begin{matrix}7-2x=0\\4+8x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

3.

<=>\(\left[{}\begin{matrix}9-7x=0\\11-3x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{7}\\x=\dfrac{11}{3}\end{matrix}\right.\)

4.

<=>\(\left[{}\begin{matrix}7-14x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=2\end{matrix}\right.\)

5. 

<=>\(\left[{}\begin{matrix}\dfrac{7}{8}-2x=0\\3x+\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{16}\\x=-\dfrac{1}{9}\end{matrix}\right.\)

6,7. ko đủ điều kiện tìm

\(\Leftrightarrow10x+6-5x^2-2x=0\)

\(\Leftrightarrow-5x^2+8x+6=0\)

\(\text{Δ}=8^2-4\cdot\left(-5\right)\cdot6=184\)>0

=>Phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{-8-2\sqrt{46}}{-10}=\dfrac{4+\sqrt{46}}{5}\\x_2=\dfrac{4-\sqrt{46}}{5}\end{matrix}\right.\)

Có: \(5x^4+10x^2+2y^6+4y^3-6=0\)

<=> \(5\left(x^4+2x^2+1\right)+2\left(y^6+2y^3+1\right)=13\)

<=> \(5\left(x^2+1\right)^2+2\left(y^3+1\right)^2=13\)

Vì x, y nguyên => \(\left(x^2+1\right)^2;\left(x^3+1\right)^2\)là số chính phương

=>  \(x^2+1=1\)

và  \(y^3+1=2\)

Khi đó: \(\hept{\begin{cases}x=0\\y=1\end{cases}}\)thử lại thỏa mãn.

\(5x^4+10x^2+2y^6+4y^3-6=0\)

\(\Leftrightarrow5x^4+10x^2+5+2y^6+4y^3+2-7-6=0\)

\(\Leftrightarrow5\left(x^4+2x^2+1\right)+2\left(y^6+2y^3+1\right)=13\)

\(\Leftrightarrow5\left(x^2+1\right)^2+2\left(y^3+1\right)^2=13\)

mà \(\left\{{}\begin{matrix}\left(x^2+1\right)^2\ge0,\forall x\inℤ\\\left(y^3+1\right)^2\ge0,\forall y\inℤ\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+1=1\\y^3+1=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2=0\\y^3=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=1\end{matrix}\right.\)

Vậy \(\left\{{}\begin{matrix}x=0\\y=1\end{matrix}\right.\) thỏa mãn yêu cầu của đề bài.

a) Ta có: 5x(12x-7)-6(10x²+3) = 0

\(\Leftrightarrow\) 60x²-35x-60x²-18 = 0

\(\Leftrightarrow\) -35x = 18

\(\Leftrightarrow\) x = \(-\dfrac{18}{35}\)

x=0 nha

x²-10x+16=0

x²-2.5x+25-9=0

x²-2.5x+25  =9

(x-5)²          =3²

x-5             =3

x                =8

Mấy bài này dễ lắm,bn làm tương tự nha(câu nào không làm theo hằng đẳng thức được thì tách

a,

Ta có

x²+5x+6=x²+6x-x+6=0

=>x(x-1)+6(x-1)=0

=>(x+6)(x-1)=0

=> x=-6 hoặc x=1

KL

b,x²-10x+16=0

=>x²-8x-2x+16=0

=>x²-2x+16-8x=0

=>x(x-2)-8(x-2)=0

=>(x-8)(x-2)=0

=>x=8 hoặc x=2

KL

c,2x²+7x+3=0

=>2x²+6x+x+3=0

=>2x(x+3)+(x+3)=0

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

=>x=-1/2 hoặc x=-3

KL

d,

x²-10x+21=0

=>x²-3x-7x+21=0

=>x(x-3)-7(x-3)=0

=>(x-7)(x-3)=0

=>x=7 hoặc x=3

KL

e,x²-2x-3=0

=>(x²-2x+1)-4=0

=>(x-1)²-2²=0

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

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

=> x=-1 hoặc x=3

KL

f,

x³-7x-6=0

=>x³+x²-x²-x-6x-6=0

=>x²(x+1)-x(x+1)-6(x+1)=0

=>(x²-x-6)(x+1)=0

=>(x²-3x+2x-6)(x+1)=0

=>[x(x+2)-3(x+2)](x+1)=0

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

=>x=-1 hoặc x=-2 hoặc x=3

KL