Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(o,x^2-9x+20=0\)
\(\Leftrightarrow x^2-4x-5x+20=0\)
\(\Leftrightarrow x\left(x-4\right)-5\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x-5=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=4\\x=5\end{cases}}\)
\(n,3x^3-3x^2-6x=0\)
\(\Leftrightarrow3x\left(x^2-x-2\right)=0\)
\(\Leftrightarrow3x\left(x^2+x-2x-2\right)=0\)
\(\Leftrightarrow3x\left[x\left(x+1\right)-2\left(x+1\right)\right]=0\)
\(\Leftrightarrow3x\left(x+1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\orbr{\begin{cases}3x=0\\x+1=0\end{cases}}\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\orbr{\begin{cases}x=0\\x=-1\end{cases}}\\x=2\end{cases}}\)
c. x^2-5x+6=0
<=> x^2-5x=-6
<=> -4x=-6
<=> x=-6/-4
vậy tập nghiệm của pt là s={-6/-4}
c. x^2-5x +6 = 0
<=> x^2 - 5x = -6
<=> - 4x = -6
<=> x= -6/-4
Mình chỉ phân tích đa thức thành nhân tử thôi , phần còn lại bạn tự tính nha keo dài lắm
A) 2x2(x+3) - x(x+3) = 0 <=> x(x - 3)(2x-1)=0
B) (2x+5)2 - (x+2)2=0 <=> (x+3)(3x+7)=0
C) (x2-2x) - (3x-6)=0 <=> (x-2)(x-3)=0
D) (2x-7)(2x-7-6x+18)=0 <=> (2x-7)(-4x+11)=0
E) (x-2)(x+1) - (x-2)(x+2)=0 <=> (x-2)*(-1)=0 <=> x-2=0
G) (2x-3)(2x+2-5x)=0 <=> (2x-3)(-3x+2)=0
H) (1-x)(5x+3+3x-7)=0 <=> (1-x)(8x-4)=0
F) (x+6)*3x=0
I) (x-3)(4x-1-5x-2)=0 <=> (x-3)(-x-3)=0
K) (x+4)(5x+8)=0
H) (x+3)(4x-9)=0
a) Ta có: \(3x^2+2x-1=0\)
\(\Leftrightarrow3x^2+3x-x-1=0\)
\(\Leftrightarrow3x\left(x+1\right)-\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy: \(S=\left\{-1;\dfrac{1}{3}\right\}\)
b) Ta có: \(x^2-5x+6=0\)
\(\Leftrightarrow x^2-2x-3x+6=0\)
\(\Leftrightarrow x\left(x-2\right)-3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
Vậy: S={2;3}
c) Ta có: \(x^2-3x+2=0\)
\(\Leftrightarrow x^2-x-2x+2=0\)
\(\Leftrightarrow x\left(x-1\right)-2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
Vậy: S={1;2}
d) Ta có: \(2x^2-6x+1=0\)
\(\Leftrightarrow2\left(x^2-3x+\dfrac{1}{3}\right)=0\)
mà \(2\ne0\)
nên \(x^2-3x+\dfrac{1}{3}=0\)
\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{23}{12}=0\)
\(\Leftrightarrow\left(x-\dfrac{3}{2}\right)^2=\dfrac{23}{12}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{2}=\dfrac{\sqrt{69}}{6}\\x-\dfrac{3}{2}=\dfrac{-\sqrt{69}}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9+\sqrt{69}}{6}\\x=\dfrac{9-\sqrt{69}}{6}\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{9+\sqrt{69}}{6};\dfrac{9-\sqrt{69}}{6}\right\}\)
e) Ta có: \(4x^2-12x+5=0\)
\(\Leftrightarrow4x^2-10x-2x+5=0\)
\(\Leftrightarrow2x\left(2x-5\right)-\left(2x-5\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{5}{2};\dfrac{1}{2}\right\}\)
$\dfrac{5x+10}{4x-8}\cdot\dfrac{4-2x}{x+2}$
$=\dfrac{5(x+2)}{4(x-2)}\cdot\dfrac{-2(x-2)}{x+2}$
$=-\dfrac{10}{4}$
$=-\dfrac{5}{2}$
$\dfrac{6x^2y^3}{8x^3y^2}$
$=\dfrac{3y}{4x}$
b)$\dfrac{x^3-x}{3x+3}$
$=\dfrac{x(x^2-1)}{3(x+1)}$
$=\dfrac{x(x-1)(x+1)}{3(x+1)}$
$=\dfrac{x(x-1)}{3}$
c)$\dfrac{x^2+3xy}{x^2-9y^2}$
$=\dfrac{x(x+3y)}{(x-3y)(x+3y)}$
$=\dfrac{x}{x-3y}$
d)$\dfrac{x^2+4x+4}{3x+6}$
$=\dfrac{(x+2)^2}{3(x+2)}$
$=\dfrac{x+2}{3}$
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
^^
$2x(x^2-4y)=2x^3-8xy$
b)$3x^2(x+3y)=3x^3+9x^2y$
c)$-\dfrac12x^2(x-3)=-\dfrac12x^3+\dfrac32x^2$
d)$(x+6)(2x-7)+x$
$=2x^2-7x+12x-42+x$
$=2x^2+6x-42$
e)$(x-5)(2x+3)+x$
$=2x^2+3x-10x-15+x$
$=2x^2-6x-15$
$6x^2+3xy=3x(2x+y)$
b)$8x^2-10xy=2x(4x-5y)$
c)$3x(x-1)-y(1-x)$
$=3x(x-1)+y(x-1)$
$=(x-1)(3x+y)$
d)$x^2-2xy+y^2-64$
$=(x-y)^2-8^2$
$=(x-y-8)(x-y+8)$
e)$2x^2+3x-5$
$=2x^2+5x-2x-5$
$=x(2x+5)-1(2x+5)$
$=(x-1)(2x+5)$
f)$16x-5x^2-3$
$=-5x^2+16x-3$
$=-5x^2+15x+x-3$
$=-5x(x-3)+(x-3)$
$=(x-3)(1-5x)$
g)$x^2-5x-6$
$=x^2-6x+x-6$
$=x(x-6)+(x-6)$
$=(x+1)(x-6)$
$2x+1=0$
$2x=-1$
$\boxed{x=-\dfrac12}$
b)$-3x-5=0$
$-3x=5$
$\boxed{x=-\dfrac53}$
c)$-6x+7=0$
$-6x=-7$
$\boxed{x=\dfrac76}$
d)$(x+6)(2x+1)=0$
$x+6=0$ hoặc $2x+1=0$
$\boxed{x=-6\text{ hoặc }x=-\dfrac12}$
e)$2x^2+7x+3=0$
$2x^2+6x+x+3=0$
$2x(x+3)+(x+3)=0$
$(2x+1)(x+3)=0$
$\boxed{x=-\dfrac12\text{ hoặc }x=-3}$
f)$(2x-3)(2x+1)=0$
$2x-3=0$ hoặc $2x+1=0$
$\boxed{x=\dfrac32\text{ hoặc }x=-\dfrac12}$
g)$2x(x-5)-x(3+2x)=26$
$2x^2-10x-3x-2x^2=26$
$-13x=26$
$\boxed{x=-2}$
h)$5x(x-1)=x-1$
$5x(x-1)-(x-1)=0$
$(x-1)(5x-1)=0$
$\boxed{x=1\text{ hoặc }x=\dfrac15}$
$x^2-6x+10=(x-3)^2+1$
Vì $(x-3)^2\ge0$
$\Rightarrow (x-3)^2+1\ge1$
$\boxed{\text{GTNN}=1}$ khi $x=3$.
b) GTNN của $2x^2-6x$$2x^2-6x=2(x^2-3x)$
$=2\left(x-\dfrac32\right)^2-\dfrac92$
Vì $\left(x-\dfrac32\right)^2\ge0$
$\Rightarrow2\left(x-\dfrac32\right)^2-\dfrac92\ge-\dfrac92$
$\boxed{\text{GTNN}=-\dfrac92}$ khi $x=\dfrac32$.
c) GTLN của $4x-x^2-5$$4x-x^2-5=-(x^2-4x+5)$
$=-(x-2)^2-1$
Vì $-(x-2)^2\le0$
$\Rightarrow -(x-2)^2-1\le-1$
$\boxed{\text{GTLN}=-1}$ khi $x=2$.
d) GTLN của $4x-x^2+3$$4x-x^2+3=-(x^2-4x-3)$
$=-(x-2)^2+7$
Vì $-(x-2)^2\le0$
$\Rightarrow -(x-2)^2+7\le7$
$\boxed{\text{GTLN}=7}$ khi $x=2$.