1)

\((x+2)(x+3)(x+4)(x+5)-24\\=[(x+2)(x+5)]\cdot[(x+3)(x+4)]-24\\=(x^2+7x+10)(x^2+7x+12)-24\)

Đặt \(x^2+7x+10=y\), khi đó biểu thức trở thành:

\(y(y+2)-24\\=y^2+2y-24\\=y^2+2y+1-25\\=(y+1)^2-5^2\\=(y+1-5)(y+1+5)\\=(y-4)(y+6)\\=(x^2+7x+10-4)(x^2+7x+10+6)\\=(x^2+7x+6)(x^2+7x+16)\)

2) Bạn xem lại đề!

\(a,(x-2)^2-25=0\\\Leftrightarrow (x-2)^2=25\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=5\\x-2=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)

\(---\)

\(b,4x(x-2)+x-2=0\\\Leftrightarrow4x(x-2)+(x-2)=0\\\Leftrightarrow(x-2)(4x+1)=0\)

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

\(---\)

\(c,4x(x-2)-x(3+4x)(?)\)

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

\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\5x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=1\end{matrix}\right.\)

\(---\)

\(e,x^2-25-(x+5)=0(sửa.đề)\\\Leftrightarrow(x^2-5^2)-(x+5)=0\\\Leftrightarrow (x-5)(x+5)-(x+5)=0\\\Leftrightarrow(x+5)(x-5-1)=0\\\Leftrightarrow(x+5)(x-6)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=6\end{matrix}\right.\)

\(---\)

\(f,5x(x-3)-x+3=0\\\Leftrightarrow5x(x-3)-(x-3)=0\\\Leftrightarrow(x-3)(5x-1)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\5x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\)

\(Toru\)

a) x⁴+x³+2x²+x+1=(x⁴+x³+x²)+(x²+x+1)=x²(x²+x+1)+(x²+x+1)=(x²+x+1)(x²+1)

b)a³+b³+c³-3abc=a³+3ab(a+b)+b³+c³ -(3ab(a+b)+3abc)=(a+b)³+c³-3ab(a+b+c)

=(a+b+c)((a+b)²-(a+b)c+c²)-3ab(a+b+c)=(a+b+c)(a²+2ab+b²-ac-ab+c²-3ab)=(a+b+c)(a²+b²+c²-ab-ac-bc)

c)Đặt x-y=a;y-z=b;z-x=c

a+b+c=x-y-z+z-x=o

đưa về như bài b

d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung

e)x²(y-z)+y²(z-x)+z²(x-y)=x²(y-z)-y²((y-z)+(x-y))+z²(x-y)

=x²(y-z)-y²(y-z)-y²(x-y)+z²(x-y)=(y-z)(x²-y²)-(x-y)(y²-z²)=(y-z)(x²-2y²+xy+xz+yz)

Ta có: a+ b + c = 0

=> a+b = - c

a^3 + b^3 + c^3 = (a+b)³ - 3a²b - 3ab² + c³

                               = ( -c)³ - 3a²b - 3ab² + c³

                               = (-c)³ +c³ -3ab( a+b)

                       =   - 3ab (-c) = 3abc ( đpcm)

2. Tìm x:

( x - 3 )² - x + 3 = 0

=> x² - 6x + 9 - x + 3 = 0

=> x² - 7x + 12 = 0

=> ( x² - 3x ) + ( 4x - 12 ) = 0

=> x.(x - 3) + 4.(x - 3) = 0

=> ( x - 3 ).( x + 4 ) = 0

=> x - 3 = 0 => x = 3

     x + 4 = 0 => x = -4

Trl:

1.

a. \(75^2+150\text{.}25+25^2\)

\(=75^2+2\text{.}75\text{.}25+25^2\)

\(=\left(75+25\right)^2\)

\(=100^2\)

\(=10000\)

b. \(2019^2-2019.19-19^2-19.1981\)

(Đề bài có sai ko vậy???)~ hoặc lak do mk ngu quá k bt lm

2. \(\left(\text{x}-3\right)^2-\text{x}+3=0\)

\(\text{x}^2-6\text{x}+9-\text{x}+3=0\)

\(\text{x}^2-7\text{x}+12=0\)

\(\text{x}^2-3\text{x}-4\text{x}+12=0\)

\(\text{x}\left(\text{x}-3\right)-4\left(\text{x}-3\right)=0\)

\(\left(\text{x}-3\right)\left(\text{x}-4\right)=0\)

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

Vậy ....

#HuyềnAnh#

Bài 1:

a) \(75^2+150.25+25^2\)

\(=75^2+2.75.25+25^2\)

\(=\left(75+25\right)^2\)

\(=100^2\)

\(=10000\)

b) \(2019^2-2019.19-19^2-19.1981\)

\(=\left(2019^2-19^2\right)-2019.19-19.1981\)

\(=\left(2019-19\right).\left(2019+19\right)-19\left(2019+1981\right)\)

\(=2000.2038-19.4000\)

\(=2000.2038-19.2.2000\)

\(=2000.2038-38.2000\)

\(=2000.\left(2038-38\right)\)

\(=2000.2000\)

\(=4000000\)

Bài 2:

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

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

\(\Rightarrow\left(x-3\right).\left(x-3-1\right)=0\)

\(\Rightarrow\left(x-3\right).\left(x-4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-3=0\\x-4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=4\end{cases}}}\)

Vậy \(x\in\left\{3;4\right\}\)

1.

\(DK:x\ge2\)

PT

\(\Leftrightarrow\left(2+x\right)\sqrt{x-2}-\left(x+2\right)\left(x-2\right)\)

\(\Leftrightarrow\left(x+2\right)\sqrt{x-2}\left(1-\sqrt{x-2}\right)=0\)

Cho này thì ok ròi nhé

2.

\(DK:x\le\frac{5}{2}\)

Xet \(x\in\left[0;\frac{5}{2}\right]\)

PT

\(\Leftrightarrow x^2-4x=5-2x\)

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

Ta co:

\(\Delta^`=\left(-1\right)^2-1.\left(-5\right)=6>0\)

\(\Rightarrow\hept{\begin{cases}x_1=1+\sqrt{6}\left(l\right)\\x_2=1-\sqrt{6}\left(l\right)\end{cases}}\)

Xet \(x\le0\)

PT

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=1\left(l\right)\\x=5\left(l\right)\end{cases}}\)

Vay PT vo nghiem 

b)

\(\left(2x-1\right)^2=25\)

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

TH1: 2x - 1 = 5

=> x = 3

TH2: 2x - 1 = -5

=> x = -2

bạn giải hô mk câu a và câu c đc ko đc thì cảm ơn bạn nhiều nhé

a) <=> x(x² - 0,25) = 0 <=> x(x-0,5)(x+0,5)=0 

<=> x = 0; 0,5; -0,5

b) (2x-1)² = 25 <=> |2x -1| = 5 :

PT có 2 nghiệm: -2 và 3.

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

PT có 2 nghiệm: -2 và 2.