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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
^^
Bài 1:
a) \(x^3-16x=x\left(x-4\right)\left(x+4\right)\)
b) \(3x^2+3y^2-6xy-12=3\left(x^2-2xy+y^2-4\right)=3\left(x-y-2\right)\left(x-y+2\right)\)
c) \(x^2+6x+5=\left(x+1\right)\left(x+5\right)\)
d) \(x^4+x^3+2x^2+x+1=\left(x^2+x+1\right)\left(x^2+1\right)\)
Bài 2:
a) Ta có: \(\left(x+6\right)^2=144\)
\(\Leftrightarrow\left[{}\begin{matrix}x+6=12\\x+6=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-18\end{matrix}\right.\)
b) Ta có: \(x^3+27+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\)
\(\Leftrightarrow x\left(x+3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=2\end{matrix}\right.\)
c) Ta có: \(2x^2-x-6=0\)
\(\Leftrightarrow2x^2-4x+3x-6=0\)
\(\Leftrightarrow2x\left(x-2\right)+3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{3}{2}\end{matrix}\right.\)
Bài 3
a) x² + 10x + 25
= x² + 2.x.5 + 5²
= (x + 5)²
b) 8x - 16 - x²
= -(x² - 8x + 16)
= -(x² - 2.x.4 + 4²)
= -(x - 4)²
c) x³ + 3x² + 3x + 1
= x³ + 3.x².1 + 3.x.1² + 1³
= (x + 1)³
d) (x + y)² - 9x²
= (x + y)² - (3x)²
= (x + y - 3x)(x + y + 3x)
= (y - 2x)(4x + y)
e) (x + 5)² - (2x - 1)²
= (x + 5 - 2x + 1)(x + 5 + 2x - 1)
= (6 - x)(3x + 4)
Bài 4
a) x² - 9 = 0
x² = 9
x = 3 hoặc x = -3
b) (x - 4)² - 36 = 0
(x - 4 - 6)(x - 4 + 6) = 0
(x - 10)(x + 2) = 0
x - 10 = 0 hoặc x + 2 = 0
*) x - 10 = 0
x = 10
*) x + 2 = 0
x = -2
Vậy x = -2; x = 10
c) x² - 10x = -25
x² - 10x + 25 = 0
(x - 5)² = 0
x - 5 = 0
x = 5
d) x² + 5x + 6 = 0
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 = -3; x = -2
a) \(x^3-16x=0\)
\(\Leftrightarrow x\left(x^2-16\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^2-16=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm4\end{cases}}\)
Vậy tập nghiệm \(S=\left\{-4;0;4\right\}\)
b) \(x^4-2x^3+10x^2-20x=0\)
\(\Leftrightarrow x^3\left(x-2\right)+10x\left(x-2\right)=0\)
\(\Leftrightarrow\left(x^3+10x\right)\left(x-2\right)=0\)
\(\Leftrightarrow x\left(x^2+10\right)\left(x-2\right)=0\)
Mà \(x^2+10>0\)nên \(\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
Vậy tập nghiệm S = { 0;2}
bài 11
a) \(x^2-xy+x\\ =x\left(x-y+1\right)\)
b)
\(x^2-2xy-4+y^2\\ =\left(x^2-2xy+y^2\right)-4\\ =\left(x-y\right)^2-4\\ =\left(x-y-2\right)\left(x-y+2\right)\)
c)
\(x^3-x^2-16x+16\\ =x^2\left(x-1\right)-16\left(x-1\right)\\ =\left(x-1\right)\left(x-4\right)\left(x+4\right)\)
bài 12
\(2x\left(x-5\right)-x\left(3+2x\right)=26\)
\(2x^2-10x-3x-2x^2=26\)
\(-13x=26\\ x=-2\)
b)
\(2\left(x+5\right)-x^2-5x=0\\ 2\left(x+5\right)-x\left(x+5\right)=0\\ \left(x+5\right)\left(2-x\right)=0\\ \left[{}\begin{matrix}x+5=0\\2-x=0\end{matrix}\right.\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
1) (x + 2)(x - 2) - (x + 3)(x + 1)
= x^2 - 4 - (x - 3)(x + 1)
= x^2 - 4 - x^2 + 2x + 3
= 2x - 1
2) a) 5(x - y) - 3x(y - x)
= 5x - 5y - 3x(y - x)
= 5x - 5y - 3xy + 3x2
b) 5x^2 - 16 + 3
= (5x^2 - x) + (-15x + 3)
= x(5x - 1) - 3(5x - 1)
= (5x - 1)(x - 3)
3) a) 2x(x + 3) + 12 - 2x^2 = 0
<=> 2x(x + 3) + 12 - 2x^2 = 0 - 12
<=> 2x(x + 3) - 2x^2 = -12
<=> x = -2
b) x^3 - 16x = 0
<=> x(x + 4)(x - 4) = 0
<=> x = 0
<=> x = 0; x = +- 4
c) (2x - 1)^2 = (x + 3)^2
<=> 4x^2 - 4x + 1 = x^2 + 6x + 9
<=> 4x^2 - 4x + 1 = x^2 + 6x + 9 - 9
<=> 4x^2 - 4x - 8 = x^2 + 6x
<=> 4x^2 - 4x - 8 = x^2 + 6x - 6x
<=> 4x^2 -10x - 8 = x^2
<=> 3x^2 - 10x - 8 = 0
<=> x = 4, x = -2/3
d) x^2 - x - 6 = 0
<=> x = -2; x = 3
\(\left(x+2\right)\left(x-2\right)-\left(x+3\right)\left(x+1\right)\)
\(=x^2-4-\left(x^2+4x+3\right)\)
\(=x^2-4-x^2-4x-3\)
\(=-4x-7\)