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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
^^
1.
\(A=\dfrac{2x-9}{\left(x-2\right)\left(x-3\right)}-\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x-2\right)\left(x-3\right)}+\dfrac{\left(2x+4\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{2x-9-\left(x^2-9\right)+\left(2x^2-8\right)}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{x^2+2x-8}{\left(x-2\right)\left(x-3\right)}=\dfrac{\left(x-2\right)\left(x+4\right)}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{x+4}{x-3}\)
b.
\(A=2\Rightarrow\dfrac{x+4}{x-3}=2\Rightarrow x+4=2\left(x-3\right)\)
\(\Rightarrow x=10\) (thỏa mãn)
2.
\(x^4+2x^2y+y^2-9=\left(x^2+y\right)^2-3^2=\left(x^2+y-3\right)\left(x^2+y+3\right)\)
a) \(x^3-5x^2+8x-4\)
\(=x^3-2x^2-3x^2+6x+2x-4\)
\(=x^2\left(x-2\right)-3x\left(x-2\right)+2\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2-3x+2\right)\)
\(=\left(x-2\right)\left(x^2-x-2x+2\right)\)
\(=\left(x-2\right)\left[x\left(x-1\right)-2\left(x-1\right)\right]\)
\(=\left(x-2\right)\left(x-1\right)\left(x-2\right)\)
b) \(A=10x^2-15x+8x-12+7\)
\(A=5x\left(2x-3\right)+4\left(2x-3\right)+7\)
\(A=\left(2x-3\right)\left(5x+4\right)+7\)
Dễ thấy \(\left(2x-3\right)\left(5x+4\right)⋮\left(2x-3\right)=B\)
Vậy để \(A⋮B\)thì \(7⋮\left(2x-3\right)\)
\(\Rightarrow2x-3\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)
\(\Rightarrow x\in\left\{2;1;5;-2\right\}\)
Vậy.......
Bài 3
a) 2x(x - 3) - x + 3 = 0
2x(x - 3) - (x - 3) = 0
(x - 3)(2x - 1) = 0
x - 3 = 0 hoặc 2x - 1 = 0
*) x - 3 = 0
x = 3
*) 2x - 1 = 0
2x = 1
x = 1/2
Vậy x = 1/2; x = 3
b) (3x - 1)(2x + 1) - (x + 1)² = 5x²
6x² + 3x - 2x - 1 - x² - 2x - 1 - 5x² = 0
(6x² - x² - 5x²) + (3x - 2x - 2x) = 0 + 1 + 1
-x = 2
x = -2
Bài 2
a) 5x² + 30y
= 5(x² + 6y)
b) x³ - 2x² - 4xy² + x
= x(x² - 2x - 4y² + 1)
= x[(x² - 2x + 1) - 4y²]
= x[(x - 1)² - (2y)²]
= x(x - 1 - 2y)(x - 1 + 2y)
Bài 3:
a: \(x^2-16=\left(x-4\right)\cdot\left(x+4\right)\)
b: \(x^2+2x+1-y^2=\left(x+1+y\right)\left(x+1-y\right)\)
c: \(=\left(x-y\right)^2-4=\left(x-y-2\right)\left(x-y+2\right)\)
Phân tích đa thức thành nhân tử
a) 2( x + 1 ) - 3y( x + 1 ) = ( x + 1 )( 2 - 3y )
b) x2 - 5x + 4 = x2 - x - 4x + 4 = x( x - 1 ) - 4( x - 1 ) = ( x - 1 )( x - 4 )
Tìm x
a) x( x - 3 ) + 7x - 21 = 0
<=> x( x - 3 ) + 7( x - 3 ) = 0
<=> ( x - 3 )( x + 7 ) = 0
<=> x - 3 = 0 hoặc x + 7 = 0
<=> x = 3 hoặc x = -7
b) ( x - 2 )2 + x( 3 - x ) = 6
<=> x2 - 4x + 4 + 3x - x2 = 6
<=> -x + 4 = 6
<=> -x = 2
<=> x = -2
\(A=\frac{x-2}{x}\)và \(B=\frac{x}{x-2}-\frac{2x}{x^2-4}\)( x ≠ 0 ; x ≠ ±3 )
a) Tại x = 23 ( tmđk ) => \(A=\frac{23-2}{23}=\frac{21}{23}\)
b) P = A.B
\(=\frac{x-2}{x}\times\left(\frac{x}{x-2}-\frac{2x}{x^2-4}\right)\)
\(=\frac{x-2}{x}\times\left(\frac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{2x}{\left(x-2\right)\left(x+2\right)}\right)\)
\(=\frac{x-2}{x}\times\frac{x^2+2x-2x}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{1}{x}\times\frac{x^2}{x+2}=\frac{x}{x+2}\)
Để P = 4 => \(\frac{x}{x+2}=4\)
=> 4( x + 2 ) = x
=> 4x + 8 - x = 0
=> 3x + 8 = 0
=> x = -8/3 ( tmđk )