\(\frac{3}{x+1}+\frac{2}{x+2}=\frac{5x+4}{x^2+3x+2}.\)ĐKXĐ: \(x\ne-1;-2\)

\(\Leftrightarrow\frac{3\left(x+2\right)}{\left(x+1\right)\left(x+2\right)}+\frac{2\left(x+1\right)}{\left(x+1\right)\left(x+2\right)}=\frac{5x+4}{\left(x+1\right)\left(x+2\right)}\)

\(\Leftrightarrow3x+6+2x+2=5x+4\)

\(\Leftrightarrow3x+2x-5x=-6-2+4\)

\(\Leftrightarrow0x=-4\)

=> PT vô nghiệm 

\(2;\frac{2}{3x-1}-\frac{15}{6x^2-x-1}=\frac{3}{2x-1}\)

\(\Leftrightarrow\frac{2\left(2x-1\right)}{\left(2x-1\right)\left(3x-1\right)}-\frac{15}{6x^2+3x-2x-1}=\frac{3\left(3x-1\right)}{\left(2x-1\right)\left(3x-1\right)}\)

\(\Leftrightarrow\frac{4x-2-15}{\left(2x-1\right)\left(3x-1\right)}=\frac{9x-3}{\left(2x-1\right)\left(3x-1\right)}\)

\(\Leftrightarrow4x-2-15=9x-3\)

\(\Leftrightarrow4x-9x=2+15-3\)

\(\Leftrightarrow-5x=14\)

.....

mấy cái này mẫu nào dài cậu phân tích ra : 

VD : câu  3 : \(3x^2-4x+1\)

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

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

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

r bắt đầu giải PHương trình :)) Mấy câu còn lại tương tự 

\(a.\left(2x-3\right)\left(4x^2+6x+9\right)-\left(2x+3\right)\left(4x^2-6x+9\right)\\ =\left(2x\right)^3-3^3-\left[\left(2x\right)^3+3^3\right]\\ =8x^3-9-\left(8x^3+9\right)\\ =8x^3-9-8x^3-9=-18\)

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

\(c.\left(3x-1\right)\left(3x+1\right)-\left(3x-2\right)^2\\ =9x^2-1-\left(9x^2-12x+4\right)\\ =9x^2-1-9x^2+12x-4\\ =12x-5\)

\(d.\left(2x-3\right)^2-\left(2x+3\right)\left(2x-3\right)\\ =\left(2x-3\right)\cdot\left[\left(2x-3\right)-\left(2x+3\right)\right]\\ =\left(2x-3\right)\cdot\left(2x-3-2x-3\right)\\ =\left(2x-3\right)\cdot\left(-6\right)\\ =-12x\cdot18\)

\(e.\left(3x-4\right)^2-\left(2x+4\right)^2\\ =9x^2-24x+16-\left(4x^2+16x+16\right)\\ =9x^2-24x+16-4x^2-16x-16\\ =5x^2-40x\)

\(f.\left(3x-5\right)^3-\left(3x+5\right)^3\\ =27x^3-135x^2+225x-125-\left(27x^3+135x^2+225x+125\right)\\ =27x^3-135x^2+225x-125-27x^3-135x^2-225x-125\\ =-270x^2-250\)

\(g.\left(2x-1\right)^2-\left(3x-1\right)^2\\ =4x^2-4x+1-\left(9x^2-6x+1\right)\\ =4x^2-4x+1-9x^2+6x-1\\ =-5x^2+2x\)

\(h.\left(x-2y\right)\left(x^2+2xy+4y^2\right)+\left(x^3-6y^3\right)\\ =x^3-8y^3+x^3-6y^3\\ =2x^3-14y^3\)

tích mình đi

ai tích mình 

mình tích lại 

thanks

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

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

\(\Rightarrow\orbr{\begin{cases}x-3=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}}\)

KL:......................

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

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

Làm  tương tự như câu a

@_@ n...h..i......ề....u  q...u.....................á!

(x²– y² + 6x + 9) : (x + y + 3) = (x² + 6x+ 9) – y² : (x + y + 3)

=(x + 3)² – y² : (x + y + 3) = (x + 3 – y) (x + 3 + y) : (x + y + 3) = (x – y + 3)

còn câu c thì sao bn

de qua

x.(2.x-1)+1/3-2/3.x=0

Bài 1

a) (x⁵ + 4x³ - 6x²) : 4x²

= 4x²(\(\dfrac{1}{4}\)x³ + x - \(\dfrac{3}{2}\)) : 4x²

= \(\dfrac{1}{4}\)x³ + x - \(\dfrac{3}{2}\)

b) (x³ - 8) : (x² + 2x + 4)

= (x - 2)(x² + 2x + 4) : (x² + 2x + 4)

= x - 2

c) (3x² - 6x) : (2 - x)

= -(6x - 3x²) : (2 - x)

= -3x(2 - x) : (2 - x)

= -3x

d) (x³ + 2x² - 2x - 1) : (x² + 3x + 1)

= [(x³ - 1) + (2x² - 2x)] : (x² + 3x + 1)

= [(x - 1)(x² + x + 1) + 2x(x - 1)] : (x² + 3x + 1)

= (x - 1)(x² + x + 1 + 2x) : (x² + 3x + 1)

= (x - 1)(x² + 3x + 1) : (x² + 3x + 1)

= x - 1

Bài 2

a) (x - 4)² - (x - 2)(x + 2) = 6

x² - 8x + 16 - (x² - 4) = 6

x² - 8x + 16 - x² + 4 = 6

-8x + 20 = 6

\(\Rightarrow\) -8x = - 14

\(\Rightarrow\) x = \(\dfrac{7}{4}\)

b) 9(x + 1)² - (3x - 2)(3x + 2) = 10

9(x² + 2x + 1) - (9x² - 4) = 10

9x² + 18x + 9 - 9x² + 4 = 10

18x + 13 = 10

\(\Rightarrow\) 18x = -3

\(\Rightarrow\) x = \(\dfrac{-1}{6}\)

Nhớ tik mik nha

không lần sau mik ko giúp đâu

AK... có j ko hiểu thì bn cứ bình luận bên dưới

Tick Là Nhấn Vào Nút Phải ko bạn!

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

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

\(\Leftrightarrow x^3-25x-x^3-8=3\)

\(\Leftrightarrow x^3-x^3-25x=8+3\)

\(\Leftrightarrow x=\frac{11}{-25}\)

Vậy x có nghiệm là \(\frac{-11}{25}.\)

\(\)

$\textbf{1)}$

$(4x-1)^2-2(4x-1)(3x-7)+(7-3x)^2$

$=(4x-1)^2-2(4x-1)(3x-7)+(3x-7)^2$

$=\left[(4x-1)-(3x-7)\right]^2$

$=(x+6)^2.$

Tại $x=44$: $(44+6)^2=50^2=2500.$

$\textbf{2)}$

$(2x-5)^2-2(2x-5)(3x-4)+(4-3x)^2$

$=(2x-5)^2-2(2x-5)(3x-4)+(3x-4)^2$

$=\left[(2x-5)-(3x-4)\right]^2$

$=(x+1)^2.$

Tại $x=24$: $(24+1)^2=625.$

1. (x + 2)(x² - 2x + 4) - (x³ + 2x²) = 5

=> x(x² - 2x + 4) + 2(x² - 2x + 4) - x³ - 2x² - 5 = 0

=> x³ - 2x² + 4x + 2x² - 4x + 8 - x³ - 2x² - 5 = 0

=> (x³ - x³) + (-2x² + 2x² - 2x²) + (4x - 4x) + (8 - 5) = 0

=> -2x² + 3 = 0

=> -2x² = -3

=> x² = 3/2

=> x = \(\pm\sqrt{\frac{3}{2}}\)

2. \(\left(x+5\right)^2-6=0\)

=> x² + 10x + 25 - 6 = 0

=> x² + 10x + 19 = 0

=> x vô nghiệm(do mình không để căn nên ghi vô nghiệm thôi nhá)

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

=> x(x² - 3x + 9) + 3(x² - 3x + 9) - x³ - 2x = 0

=> x³ - 3x² + 9x + 3x² - 9x + 27 - x³ - 2x = 0

=> (x³ - x³) + (-3x² + 3x²) + (9x - 9x - 2x) + 27 = 0

=> -2x + 27 = 0

=> -2x = -27

=> x = 27/2

4. \(\left(x-2\right)^3-x^3+6x^2=7\)

=> x³ - 6x²  + 12x - 8 - x³ + 6x² = 7

=> (x³ - x³) + (-6x² + 6x²) + 12x - 8 = 7

=> 12x - 8 = 7

=> 12x = 15

=> x = 5/4

5. \(3\left(x-2\right)^2+9\left(x-1\right)-3\left(x^2+x-3\right)=12\)

=> 3x² - 12x + 12 + 9x - 9 - 3x² - 3x + 9 = 12

=> (3x² - 3x²) + (-12x + 9x - 3x) + (12 - 9 + 9) = 12

=> -6x + 12 = 12

=> -6x = 0

=> x = 0

6. \(\left(4x+3\right)^2-\left(4x-3\right)^2-5x-2=0\)

=> 48x - 5x - 2 = 0

=> 43x - 2 = 0

=> 43x = 2

=> x = 2/43

Còn bài cuối tự làm :>

Anh Sang làm cầu kì quá ;-;

1. ( x + 2 )( x² - 2x + 4 ) - ( x³ + 2x² ) = 5

<=> x³ + 8 - x³ - 2x² = 5

<=> 8 - 2x² = 5

<=> 2x² = 3

<=> x² = 3/2

<=> \(x^2=\left(\pm\sqrt{\frac{3}{2}}\right)^2\)

<=> \(x=\pm\sqrt{\frac{3}{2}}\)

2. ( x + 5 )² - 6 = 0

<=> ( x + 5 )² - ( √6 )² = 0

<=> ( x + 5 - √6 )( x + 5 + √6 ) = 0

<=> \(\orbr{\begin{cases}x+5-\sqrt{6}=0\\x+5+\sqrt{6}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\sqrt{6}-5\\x=-\sqrt{6}-5\end{cases}}\)

3. ( x + 3 )( x² - 3x + 9 ) - x³ = 2x

<=> x³ + 27 - x³ = 2x

<=> 27 = 2x

<=> x = 27/2

4. ( x - 2 )³ - x³ + 6x² = 7

<=> x³ - 6x² + 12x - 8 - x³ + 6x² = 7

<=> 12x - 8 = 7

<=> 12x = 15

<=> x = 15/12 = 5/4

5. 3( x - 2 )² + 9( x - 1 ) - 3( x² + x - 3 ) = 12

<=> 3( x² - 4x + 4 ) + 9x - 9 - 3x² - 3x + 9 = 12

<=> 3x² - 12x + 12 + 6x - 3x² = 12

<=> -6x + 12 = 12

<=> -6x = 0

<=> x = 0

6. ( 4x + 3 )² - ( 4x - 3 )² - 5x - 2 = 0

<=> 16x² + 24x + 9 - ( 16x² - 24x + 9 ) - 5x - 2 = 0

<=> 16x² + 24x + 9 - 16x² + 24x - 9 - 5x - 2 = 0

<=> 43x - 2 = 0

<=> 43x = 2

<=> x = 2/43

7, ( 4x + 7 )( 2 - 3x ) - ( 6x + 2 )( 5 - 2x ) = 0

<=> -12x² - 13x + 14 - ( -12x² + 26x + 10 ) = 0

<=> -12x² - 13x + 14 + 12x² - 26x - 10 = 0

<=> -39x + 4 = 0

<=> -39x = -4

<=> x = 4/39

Tương tự mấy phần kia 

\(A=\frac{x+3}{x-2}+\frac{x+2}{3-x}+\frac{x+2}{x^2-5x+6}\)

\(=\frac{x+3}{x-2}-\frac{x+2}{x-3}+\frac{x+2}{\left(x-2\right)\left(x-3\right)}\)

\(=\frac{\left(x+3\right)\left(x-3\right)}{\left(x-2\right)\left(x-3\right)}-\frac{\left(x+2\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}+\frac{x+2}{\left(x-2\right)\left(x-3\right)}\)

\(=\frac{x^2-9-x^2+4+x+2}{\left(x-2\right)\left(x-3\right)}=\frac{-3+x}{\left(x-2\right)\left(x-3\right)}=\frac{-1}{x-2}\)