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a: \(=\dfrac{6x^3+13x^2-5x}{2x+5}=\dfrac{6x^3+15x^2-2x^2-5x}{2x+5}=3x^2-x\)
b: \(=\dfrac{x^4-6x^3+12x^2-14x+3}{x^2-4x+1}\)
\(=\dfrac{x^4-4x^3+x^2-2x^3+8x^2-2x+3x^2-12x+3}{x^2-4x+1}\)
\(=x^2-2x+3\)
d: \(=\dfrac{\left(x+y\right)^2}{x+y}=x+y\)
Rút gọn hết ta được :
a/ 41x - 17 = -21
=> 41x = -4 => x = 4/41
b/ 34x - 17 = 0
=> 34x = 17
=> x = 17/34 = 1/2
c/ 19x + 56 = 52
=> 19x = -4
=> x = -4/19
d/ 20x2 - 16x - 34 = 10x2 + 3x - 34
=> 10x2 - 19x = 0
=> x(10x - 19) = 0
=> x = 0
hoặc 10x - 19 = 0 => 10x = 19 => x = 19/10
Vậy x = 0 ; x = 19/10
Rút gọn hết ta được :
a/ 41x - 17 = -21
=> 41x = -4 => x = 4/41
b/ 34x - 17 = 0
=> 34x = 17
=> x = 17/34 = 1/2
c/ 19x + 56 = 52
=> 19x = -4
=> x = -4/19
d/ 20x 2 - 16x - 34 = 10x 2 + 3x - 34
=> 10x 2 - 19x = 0
=> x(10x - 19) = 0
=> x = 0 hoặc 10x - 19 = 0
=> 10x = 19
=> x = 19/10
Vậy x = 0 ; x = 19/10
a: \(\Leftrightarrow2x^2-6x+x-3=5x+4-2\left(x^2-4\right)\)
\(\Leftrightarrow2x^2-5x-3=5x+4-2x^2+8\)
\(\Leftrightarrow4x^2-10x-9=0\)
\(\text{Δ}=\left(-10\right)^2-4\cdot4\cdot\left(-9\right)=100+144=244>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{10-\sqrt{244}}{8}=\dfrac{5-\sqrt{61}}{4}\\x_2=\dfrac{5+\sqrt{61}}{4}\end{matrix}\right.\)
b: \(\Leftrightarrow3x-4x^2-4x-1=4x^2-\left(2x^2+2x-x-1\right)\)
\(\Leftrightarrow-4x^2-x-1-4x^2+2x^2+x-1=0\)
\(\Leftrightarrow-6x^2-2=0\)
hay \(x\in\varnothing\)
c: \(\dfrac{3x-1}{x-1}-\dfrac{2x+5}{x+3}=1-\dfrac{4}{x^2+2x-3}\)
\(\Leftrightarrow\left(3x-1\right)\left(x+3\right)-\left(2x+5\right)\left(x-1\right)=x^2+2x-3-4\)
\(\Leftrightarrow3x^2+9x-x-3-2x^2+2x-5x+5=x^2+2x-7\)
=>5x+2=2x-7
=>3x=-9
hay x=-3(loại)
\(\frac{x^4+x^3+6x^2+5x+5}{x^2+x+1}=\frac{x^4+x^3+x^2+5x^2+5x+5}{x^2+x+1}=\frac{x^2\left(x^2+x+1\right)+5\left(x^2+x+1\right)}{\left(x^2+x+1\right)}=\frac{\left(x^2+x+1\right)\left(x^2+5\right)}{x^2+x+1}=x^2+5\)
\(\frac{x^4+x^3+2x^2+x+1}{x^2+x+1}=\frac{x^4+x^3+x^2+x^2+x+1}{x^2+x+1}=\frac{x^2\left(x^2+x+1\right)+\left(x^2+x+1\right)}{x^2+x+1}=\frac{\left(x^2+x+1\right)\left(x^2+1\right)}{x^2+x+1}=x^2+1\)
\(a,\left(6x+1\right)\left(x+2\right)-2x\left(3x-5\right)\)
\(=6x^2+12x+x+2-6x^2+10x\)
\(=23x+2\)
a) (6x + 1)(x + 2) - 2x(3x - 5)
= 6x2 + 12x + x + 2 - 6x2 + 10x
= (6x2 - 6x2) + (12x + x + 10x) + 2
= 23x + 2
b) (2x - 1)2 - (2x - 3)(2x + 3)
= 4x2 - 4x + 1 - 4x2 + 9
= (4x2 - 4x2) - 4x + (1 + 9)
= -4x + 10
c) (2x - 3)3 - (3x + 1)(5 - 4x) - 16x2
= 8x3 - 36x2 + 54x - 15x + 12x2 - 5 + 4x - 16x2
= 8x3 - (36x2 - 12x2 + 16x2) + (54x - 15x + 4x) - 5
= 8x3 - 40x2 + 43x - 5
d) (3x + 2) - (x - 5) - x(3x - 13)
= 3x + 2 - x + 5 - 3x2 + 13x
= (3x - x + 13x) + (2 + 5) - 3x2
= 15x + 7 - 3x2
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
^^
a) Ta có: \(\frac{x^3-3x^2+x-3}{x-3}\)
\(=\frac{x^2\left(x-3\right)+\left(x-3\right)}{\left(x-3\right)}=\frac{\left(x-3\right)\left(x^2+1\right)}{x-3}=x^2+1\)
b) Ta có: \(\frac{x^2+2x+x^2-4}{x+2}\)
\(=\frac{x\left(x+2\right)+\left(x+2\right)\left(x-2\right)}{x+2}=\frac{\left(x+2\right)\left(x+x-2\right)}{x+2}=2x-2\)
c) Ta có: \(\frac{2x^3-5x^2+6x-15}{2x-5}\)
\(=\frac{x^2\left(2x-5\right)+3\left(2x-5\right)}{2x-5}=\frac{\left(2x-5\right)\left(x^2+3\right)}{2x-5}=x^2+3\)
cái này chia dọc hay ngang