Bài 5 : 

f, bạn xem lại đề hay là tìm x chứa tham số a ? 

g, \(x^2+3x-\left(2x+6\right)=0\Leftrightarrow x\left(x+3\right)-2\left(x+3\right)=0\)

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

h, \(5x+20-x^2-4x=0\Leftrightarrow5\left(x+4\right)-x\left(x+4\right)=0\)

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

m, \(x^3-5x^2-x+5=0\Leftrightarrow x^2\left(x-5\right)-\left(x-5\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-5\right)=0\Leftrightarrow x=\pm1;x=5\)

n, \(x\left(x-3\right)-7x+21=0\Leftrightarrow x\left(x-3\right)-7\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-7\right)\left(x-3\right)=0\Leftrightarrow x=3;x=7\)

x=7 nha

a) \(2.\left(x+5\right)-x^2-5x=0\)

\(\Leftrightarrow2.\left(x+5\right)-x.\left(x+5\right)=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x+5=0\\2-x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=2\end{cases}}\)

Vậy \(S=\left\{-5,2\right\}\)

b) \(x^3-5x^2-4x+20=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\x^2-4=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=5\\x=\pm2\end{cases}}\)

Vậy \(S=\left\{5,\pm2\right\}\)

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

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

\(\Leftrightarrow\left(x-4\right)\left(3x+2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\3x+2=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=4\\x=-\frac{3}{2}\end{cases}}\)

Vậy \(S=\left\{4,-\frac{3}{2}\right\}\)

. Ai đó giúp tôi đi mà ._.

bài khó quá bạn ạ

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

\(< =>2x+10-x^2-5x=0\)

\(< =>-x^2-3x+10=0\)

\(< =>-\left(x^2+3x+\frac{9}{4}\right)+\frac{49}{4}=0\)

\(< =>-\left(x+\frac{3}{2}\right)^2=-\frac{49}{4}\)

\(< =>\left(x+\frac{3}{2}\right)^2=\frac{49}{4}< =>\orbr{\begin{cases}x+\frac{3}{2}=\sqrt{\frac{49}{4}}\\x+\frac{3}{2}=-\sqrt{\frac{49}{4}}\end{cases}}\)

\(< =>\orbr{\begin{cases}x=\frac{7}{2}-\frac{3}{2}=\frac{4}{2}=2\\x=-\frac{7}{2}-\frac{3}{2}=-\frac{10}{2}=-5\end{cases}}\)

b, Đật x = y+5/3 khi đó phương trình trở thành 

\(y^3-\frac{37}{3}y+\frac{476}{27}=0\)

Đặt \(y=u+v\)sao cho uv=37/9 thế vào ta được phương trình mới sau ta được

\(u^3+v^3+\left(3uv-\frac{37}{3}\right)\left(u+v\right)+\frac{426}{27}=0\)

Khi đó ta có hệ sau : \(\hept{\begin{cases}u^3+v^3=-\frac{426}{27}\\u^3v^3=\frac{50653}{729}\end{cases}}\)

Theo Vi ét u^3 và v^3 là 2 nghiệm của pt \(x^2-\frac{426}{27}x+\frac{50653}{729}=0\)

Đến đây delta phát rồi tìm ngược lại là xong :))))

mình dùng cardano nhưng làm trong nháp xong gửi nên chắc chắc bạn sẽ không hiểu được :V

làm luôn câu cuối nhé ^^

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

\(\Leftrightarrow\left(4x^2-4x+1\right)-\left(x^2+6x+9\right)=0\)

\(\Leftrightarrow4x^2-4x+1-x^2-6x-9=0\)

\(\Leftrightarrow3x^2-10x-8=0\)

\(\Leftrightarrow3\left(x^2-\frac{10}{3}x+\frac{25}{9}\right)-\frac{147}{9}=0\)

\(\Leftrightarrow3\left(x-\frac{5}{3}\right)^2=\frac{147}{9}\Leftrightarrow\left(x-\frac{5}{3}\right)^2=\frac{147}{27}\)

\(\Leftrightarrow\orbr{\begin{cases}x-\frac{5}{3}=\sqrt{\frac{147}{27}}\\x-\frac{5}{3}=-\sqrt{\frac{147}{27}}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\sqrt{\frac{147}{27}}+\frac{5}{3}\\x=-\sqrt{\frac{147}{27}}+\frac{5}{3}\end{cases}}\)

bạn có thể giải rõ giúp mình phần b được không ?

a)x²-20-x=0

<=>(x²-5x)+(4x-20)=0

<=>x(x-5)+4(x-5)=0

<=>(x-5)(x+4)=0

<=>x-5=0 hoặc x+4=0

<=>x=5 hoặc x=-4

b)(2x+3)²-(4x²-9)=0

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

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

<=>(2x+3).6=0

<=>2x+3=0

<=>2x=-3

<=>x=-1,5

c)(2x²+5x+3):(x+1)=4x-5

<=>2x²+5x+3=(4x-5)(x+1)

<=>2x²+5x+3=4x²-x-5

<=>4x²-x-5-2x²-5x-3=0

<=>2x²-6x-8=0

<=>x²-3x-4=0

<=>(x²-4x)+(x-4)=0

<=>x(x-4)+(x-4)=0

<=>(x-4)(x+1)=0

<=>x+1=0 hoặc x-4=0

<=>x=-1 hoặc x=4

Bài 1

a) 5x²y - 20xy²

= 5xy(x - 4y)

b) 1 - 8x + 16x² - y²

= (1 - 8x + 16x²) - y²

= (1 - 4x)² - y²

= (1 - 4x - y)(1 - 4x + y)

c) 4x - 4 - x²

= -(x² - 4x + 4)

= -(x - 2)²

d) x³ - 2x² + x - xy²

= x(x² - 2x + 1 - y²)

= x[(x² - 2x+ 1) - y²]

= x[(x - 1)² - y²]

= x(x - 1 - y)(x - 1 + y)

= x(x - y - 1)(x + y - 1)

e) 27 - 3x²

= 3(9 - x²)

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

f) 2x² + 4x + 2 - 2y²

= 2(x² + 2x + 1 - y²)

= 2[(x² + 2x + 1) - y²]

= 2[(x + 1)² - y²]

= 2(x + 1 - y)(x + 1 + y)

= 2(x - y + 1)(x + y + 1)

Bài 2:

a: \(x^2\left(x-2023\right)+x-2023=0\)

=>\(\left(x-2023\right)\left(x^2+1\right)=0\)

mà \(x^2+1>=1>0\forall x\)

nên x-2023=0

=>x=2023

b: 

ĐKXĐ: x<>0

\(-x\left(x-4\right)+\left(2x^3-4x^2-9x\right):x=0\)

=>\(-x\left(x-4\right)+2x^2-4x-9=0\)

=>\(-x^2+4x+2x^2-4x-9=0\)

=>\(x^2-9=0\)

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

=>\(\left[{}\begin{matrix}x-3=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

c: \(x^2+2x-3x-6=0\)

=>\(\left(x^2+2x\right)-\left(3x+6\right)=0\)

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

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

=>\(\left[{}\begin{matrix}x+2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

d: 3x(x-10)-2x+20=0

=>\(3x\left(x-10\right)-\left(2x-20\right)=0\)

=>\(3x\left(x-10\right)-2\left(x-10\right)=0\)

=>\(\left(x-10\right)\left(3x-2\right)=0\)

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

Câu 1:

a: \(5x^2y-20xy^2\)

\(=5xy\cdot x-5xy\cdot4y\)

\(=5xy\left(x-4y\right)\)

b: \(1-8x+16x^2-y^2\)

\(=\left(16x^2-8x+1\right)-y^2\)

\(=\left(4x-1\right)^2-y^2\)

\(=\left(4x-1-y\right)\left(4x-1+y\right)\)

c: \(4x-4-x^2\)

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

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

d: \(x^3-2x^2+x-xy^2\)

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

\(=x\left[\left(x^2-2x+1\right)-y^2\right]\)

\(=x\left[\left(x-1\right)^2-y^2\right]\)

\(=x\left(x-1-y\right)\left(x-1+y\right)\)

e: \(27-3x^2\)

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

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

f: \(2x^2+4x+2-2y^2\)

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

\(=2\left[\left(x^2+2x+1\right)-y^2\right]\)

\(=2\left[\left(x+1\right)^2-y^2\right]\)

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

Bài 2

a) x²(x - 2023) - 2023 + x = 0

x²(x - 2023) - (x - 2023) = 0

(x - 2023)(x² - 1) = 0

x - 2023 = 0 hoặc x² - 1 = 0

*) x - 2023 = 0

x = 2023

*) x² - 1 = 0

x² = 1

x = 1 hoặc x = -1

Vậy x = -1; x = 1; x = 2023

b) -x(x - 4) + (2x³ - 4x² - 9x) : x = 0

-x² + 4x + 2x² - 4x - 9 = 0

x² - 9 = 0

x² = 9

x = 3 hoặc x = -3

Vậy x = 3; x = -3

c) x² + 2x - 3x - 6 = 0

(x² + 2x) - (3x + 6) = 0

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

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

x + 2 = 0 hoặc x - 3 = 0

*) x + 2 = 0

x = -2

*) x - 3 = 0

x = 3

Vậy x = -2; x = 3

d) 3x(x - 10) - 2x + 20 = 0

3x(x - 10) - (2x - 20) = 0

3x(x - 10) - 2(x - 10) = 0

(x - 10)(3x - 2) = 0

x - 10 = 0 hoặc 3x - 2 = 0

*) x - 10 = 0

x = 10

*) 3x - 2 = 0

3x = 2

x = 2/3

Vậy x = 2/3; x = 10

Bài 1 : 

a, \(\left(x-3\right)^2-4=0\Leftrightarrow\left(x-3\right)^2=4\Leftrightarrow\left(x-3\right)^2=\left(\pm2\right)^2\)

TH1 : \(x-3=2\Leftrightarrow x=5\)

TH2 : \(x-3=-2\Leftrightarrow x=1\)

b, \(x^2-2x=24\Leftrightarrow x^2-2x-24=0\)

\(\Leftrightarrow\left(x-6\right)\left(x+4\right)=0\)

TH1 : \(x-6=0\Leftrightarrow x=6\)

TH2 : \(x+4=0\Leftrightarrow x=-4\)

c, \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+2\right)\left(x-2\right)=0\)

\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5\left(x^2-4\right)=0\)

\(\Leftrightarrow2x+30=0\Leftrightarrow x=-15\)

d, tương tự 

Bài 2 :

 \(x^2+2xy+y^2-6x-6y-5=\left(x+y\right)^2-6\left(x+y\right)-5\)

Thay x + y = -9 ta có : 

\(\left(-9\right)^2-6\left(-9\right)-5=130\)

Bài 1.

a) ( x - 3 )² - 4 = 0

<=> ( x - 3 )² - 2² = 0

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

<=> ( x - 5 )( x - 1 ) = 0

<=> \(\orbr{\begin{cases}x-5=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=5\\x=1\end{cases}}\)

b) x² - 2x = 24

<=> x² - 2x - 24 = 0

<=> x² + 4x - 6x - 24 = 0

<=> x( x + 4 ) - 6( x + 4 ) = 0

<=> ( x + 4 )( x - 6 ) = 0

<=> \(\orbr{\begin{cases}x+4=0\\x-6=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-4\\x=6\end{cases}}\)

c) ( 2x - 1 )² + ( x + 3 )² - 5( x + 2 )( x - 2 ) = 0

<=> 4x² - 4x + 1 + x² + 6x + 9 - 5( x² - 4 ) = 0

<=> 5x² + 2x + 10 - 5x² + 20 = 0

<=> 2x + 30 = 0

<=> 2x = -30

<=> x = -15

Bài 2.

D = x² + 2xy + y² - 6x - 6y - 5

= [ ( x² + 2xy + y² ) - 2x - 2y + 1 ] - 4x - 4y - 6

= [ ( x + y )² - 2( x + y ) + 1 ] - 4( x + y ) - 6

= ( x + y - 1 )² - 4( x + y ) - 6

Với x + y = -9

D = ( -9 - 1 )² - 4.(-9) - 6

    = 100 + 36 - 6

    = 130

Bài 3.

4x² + y² - 4x + 10y + 26 = 0

<=> ( 4x² - 4x + 1 ) + ( y² + 10y + 25 ) = 0

<=> ( 2x - 1 )² + ( y + 5 )² = 0

<=> \(\hept{\begin{cases}2x-1=0\\y+5=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=-5\end{cases}}\)

Bài 1:

a) \(x^2-5x+1=0\)

\(\Leftrightarrow\left(x^2-5x+\frac{25}{4}\right)-\frac{21}{4}=0\)

\(\Leftrightarrow\left(x-\frac{5}{2}\right)^2-\frac{\left(\sqrt{21}\right)^2}{2^2}=0\)

\(\Leftrightarrow\left(x-\frac{5+\sqrt{21}}{2}\right)\left(x+\frac{\sqrt{21}-5}{2}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-\frac{5+\sqrt{21}}{2}=0\\x+\frac{\sqrt{21}-5}{2}=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{5+\sqrt{21}}{2}\\x=\frac{5-\sqrt{21}}{2}\end{cases}}\)

b) \(3x^2-12x-1=0\)

\(\Leftrightarrow3\left(x^2-4x+4\right)-13=0\)

\(\Leftrightarrow\left(x-2\right)^2-\left(\sqrt{\frac{13}{3}}\right)^2=0\)

\(\Leftrightarrow\left(x-2-\sqrt{\frac{13}{3}}\right)\left(x-2+\sqrt{\frac{13}{3}}\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=2+\sqrt{\frac{13}{3}}\\x=2-\sqrt{\frac{13}{3}}\end{cases}}\)

Bài 2:

a) \(A=\frac{1}{4}x^2-x+1=\left(\frac{1}{2}x-1\right)^2\ge0\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(\left(\frac{1}{2}x-1\right)^2=0\Rightarrow\frac{1}{2}x=1\Rightarrow x=2\)

Vậy Min(A) = 0 khi x = 2

b) \(B=3x^2-4x-2=3\left(x^2-\frac{4}{3}x+\frac{4}{9}\right)-\frac{10}{3}=3\left(x-\frac{2}{3}\right)^2-\frac{10}{3}\ge-\frac{10}{3}\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(3\left(x-\frac{2}{3}\right)^2=0\Rightarrow x=\frac{2}{3}\)

Vậy \(Min\left(B\right)=-\frac{10}{3}\Leftrightarrow x=\frac{2}{3}\)