Giải như sau.

(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y

⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn ! 

\(\left(x+6\right)\left(2x+1\right)=0\)

<=>  \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)

<=>  \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)

Vậy....

hk tốt

^^

Trần văn ổi ()

đù khó thế

tl j z mấy chế , k câu dc đâu :))

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

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

<=> \(\orbr{\begin{cases}3x+2=0\\x-1=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=-\frac{2}{3}\\x=1\end{cases}}\)

Vậy S = {-2/3; 1}

b) x² - 1 - (x + 5)(2 - x) = 0

<=> x² - 1 - 2x + x² - 10 + 5x = 0

<=> 2x² + 3x - 11 = 0

<=> 2(x² + 3/2x + 9/16 - 97/16) = 0

<=> (x + 3/4)² - 97/16 = 0

<=> \(\orbr{\begin{cases}x+\frac{3}{4}=\frac{\sqrt{97}}{4}\\x+\frac{3}{4}=-\frac{\sqrt{97}}{4}\end{cases}}\)

<=> \(\orbr{\begin{cases}x=\frac{\sqrt{97}-3}{4}\\x=-\frac{\sqrt{97}-3}{4}\end{cases}}\)

Vậy S = {\(\frac{\sqrt{97}-3}{4}\); \(-\frac{\sqrt{97}-3}{4}\)

d) x(2x - 3) - 4x + 6 = 0

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

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

<=> \(\orbr{\begin{cases}x-2=0\\2x-3=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=2\\x=\frac{3}{2}\end{cases}}\)

Vậy  S = {2; 3/2}

e)  x³ - 1 = x(x - 1)

<=> (x - 1)(x² + x + 1) - x(x - 1) = 0

<=> (x - 1)(x² + x +  1 - x) = 0

<=> (x - 1)(x² + 1) = 0

<=> x - 1 = 0

<=> x = 1

Vậy S = {1}

f) (2x - 5)² - x² - 4x - 4 = 0

<=> (2x - 5)² - (x + 2)² = 0

<=> (2x - 5 - x - 2)(2x - 5 + x + 2) = 0

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

<=> \(\orbr{\begin{cases}x-7=0\\3x-3=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=7\\x=1\end{cases}}\)

Vậy S = {7; 1}

h) (x - 2)(x² + 3x - 2) - x³ + 8 = 0

<=> (x - 2)(x² + 3x - 2) - (x- 2)(x² + 2x + 4) = 0

<=> (x - 2)(x² + 3x - 2 - x² - 2x - 4) = 0

<=> (x - 2)(x - 6) = 0

<=> \(\orbr{\begin{cases}x-2=0\\x-6=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=2\\x=6\end{cases}}\)

Vậy S = {2; 6}

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

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

\(2x-2=0\)

\(2x=2\)

\(x=1\)

_{Sorry mình nhầm câu a}

a) (2x - 1)² + (x + 3)² - 5(x + 7)(x - 7) = 0

b) (x + 2)(x² - 2x + 4) - x(x² + 2) = 15

c) (x + 3)³ - x(3x + 1)² + (2x - 1)(4x² - 2x + 1) = 28

d) (x² - 1)³ - (x⁴ + x² + 1)(x² - 1) = 0

Giải:

a) (2x - 1)² + (x + 3)² - 5(x + 7)(x - 7) = 0

\(\Leftrightarrow\) 4x² - 4x + 1 + x² + 6x + 9 - 5(x² - 49) = 0

\(\Leftrightarrow\) 4x² - 4x + 1 + x² + 6x + 9 - 5x² + 245 = 0

\(\Leftrightarrow\) 2x + 255 = 0

\(\Leftrightarrow\) 2x = - 255

\(\Leftrightarrow\) x = - 255 : 2

\(\Leftrightarrow\) x = \(-\frac{255}{2}\)

Vậy x = \(-\frac{255}{2}\)

b) (x + 2)(x² - 2x + 4) - x(x² + 2) = 15

\(\Leftrightarrow\) x³ + 8 - x³ - 2x = 15

\(\Leftrightarrow\) 8 - 2x = 15

\(\Leftrightarrow\) 2x = 8 - 1

\(\Leftrightarrow\) 2x = - 7

\(\Leftrightarrow\) x = - 7 : 2

\(\Leftrightarrow\) x = \(-\frac{7}{2}\)

Vậy x = \(-\frac{7}{2}\)

c) (x + 3)³ - x(3x + 1)² + (2x - 1)(4x² - 2x + 1) = 28

\(\Leftrightarrow\) x³ + 6x² + 27x + 27 - x(9x² + 6x + 1) + 8x³ - 1 = 28

\(\Leftrightarrow\) x³ + 6x² + 27x + 27 - 9x³ - 6x² - x + 8x³ - 1 = 28

\(\Leftrightarrow\) 26x + 26 = 28

\(\Leftrightarrow\) 26x = 28 - 26

\(\Leftrightarrow\) 26x = 2

\(\Leftrightarrow\) x = 2 : 26

\(\Leftrightarrow\) x = \(\frac{1}{13}\)

Vậy x = \(\frac{1}{13}\)

d) (x² - 1)³ - (x⁴ + x² + 1)(x² - 1) = 0

\(\Leftrightarrow\) x⁶ - 2x² + 1 - (x⁶ - 1) = 0

\(\Leftrightarrow\) x⁶ - 2x² + 1 - x⁶ + 1 = 0

\(\Leftrightarrow\) -2x² + 2 = 0

\(\Leftrightarrow\) -2x² = - 2

\(\Leftrightarrow\) x² = - 2 : (- 2)

\(\Leftrightarrow\) x² = 1

\(\Leftrightarrow\) x = 1 hoặc x = - 1

Vậy x \(\in\) {1; - 1}

cơ may sao nay gặp được Dân Chơi Đất Bắc luôn nè ::)))

Bài 1:

a) (3x - 2)(4x + 5) = 0

<=> 3x - 2 = 0 hoặc 4x + 5 = 0

<=> 3x = 2 hoặc 4x = -5

<=> x = 2/3 hoặc x = -5/4

b) (2,3x - 6,9)(0,1x + 2) = 0

<=> 2,3x - 6,9 = 0 hoặc 0,1x + 2 = 0

<=> 2,3x = 6,9 hoặc 0,1x = -2

<=> x = 3 hoặc x = -20

c) (4x + 2)(x^2 + 1) = 0

<=> 4x + 2 = 0 hoặc x^2 + 1 # 0

<=> 4x = -2

<=> x = -2/4 = -1/2

d) (2x + 7)(x - 5)(5x + 1) = 0

<=> 2x + 7 = 0 hoặc x - 5 = 0 hoặc 5x + 1 = 0

<=> 2x = -7 hoặc x = 5 hoặc 5x = -1

<=> x = -7/2 hoặc x = 5 hoặc x = -1/5

bài 2:

a, (3x+2)(x^2-1)=(9x^2-4)(x+1)

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

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

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

b, x(x+3)(x-3)-(x-2)(x^2-2x+4)=0

x(x^2-9)-(x^3+8)=0

x^3-9x-x^3-8=0

-9x-8=0

tự tìm x nha

a) \(\left(x-2\right)\left(x^2+2x+7\right)+2\left(x^2-4\right)-5\left(x-2\right)=0\)

\(\Rightarrow\left(x-2\right)\left(x^2+2x+7\right)+2\left(x-2\right)\left(x+2\right)-5\left(x-2\right)=0\)

\(\Rightarrow\left(x-2\right)\left[x^2+2x+7+2\left(x+2\right)-5\right]=0\)

\(\Rightarrow\left(x-2\right)\left(x^2+2x+7+2x+4-5\right)=0\)

\(\Rightarrow\left(x-2\right)\left(x^2+4x+6\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-2=0\\x^2+4x+6=0\end{matrix}\right.\)

Ta có:

\(x^2+4x+6\)

\(=x^2+2.x.2+4+2\)

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

Vì \(\left(x+2\right)^2\ge0\) với mọi x

\(\Rightarrow\left(x+2\right)^2+2\ge2\) với mọi x

\(\Rightarrow x^2+4x+6\) vô nghiệm

\(\Rightarrow x-2=0\)

\(\Rightarrow x=2\)

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

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

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

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

\(\Rightarrow x\left[2\left(x+3\right)x-3\right]=0\)

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

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

Ta có:

\(2x^2+6x-3\)

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

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

\(=2\left(x+\dfrac{3}{2}\right)^2-\dfrac{15}{2}\)

Vì \(2\left(x+\dfrac{3}{2}\right)^2\ge0\) với mọi x

\(\Rightarrow2\left(x+\dfrac{3}{2}\right)^2-\dfrac{15}{2}\ge-\dfrac{15}{2}\) với mọi x

\(\Rightarrow2x^2+6x-3\) vô nghiệm

\(\Rightarrow x=0\)

a)

\(A=3\left(x-y\right)^2-2\left(x+y\right)^2-\left(x-y\right)\left(x+y\right)\)\(2A=\left[\left(x-y\right)-\left(x+y\right)\right]^2+5\left(x-y\right)^2-5\left(x+y\right)^2\)

\(2A=4y^2+5\left[\left(x-y\right)-\left(x+y\right)\right]\left[\left(x-y\right)+\left(x+y\right)\right]\)\(2A=4y^2+5\left[-2y\right]\left[2x\right]=4y^2-20xy=4y\left(y-5x\right)\\ \)\(A=2y\left(y-5x\right)\)