Tìm x
a) x^3 - 16x = 0
b) x^4 - 2x^3 + 10x^2 - 20x = 0
c) (2x - 3 )^2 = (x+5)^2
d) x^2(x-1) - 4x^2 + 8x - 4 = 0
e) x^2 + 4x + 3 = 0
f) x^3 - x^2 = 4x^2 - 8x + 4
g) 2(x+3) - x^2 - 3x = 0
tìm x: a)x^4-2x^3+5x^2-10x=0
b)(3x+5)^2=(2x-2)^2
. c)x^3–2x^2+x=0
. d)x^2(x-1)-4x^2+8x-4=0
\(a,x^4-2x^3+5x^2-10x=0\\ \Leftrightarrow x^3\left(x-2\right)+5x\left(x-2\right)=0\\ \Leftrightarrow x\left(x^2+5\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x^2+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x\in\varnothing\left(x^2+5>0\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
\(b,\left(3x+5\right)^2=\left(2x-2\right)^2\\ \Leftrightarrow\left(3x+5\right)^2-\left(2x-2\right)^2=0\\ \Leftrightarrow\left(3x+5+2x-2\right)\left(3x+5-2x+2\right)=0\\ \Leftrightarrow\left(5x+3\right)\left(x+7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{5}\\x=-7\end{matrix}\right.\)
\(c,x^3-2x^2+x=0\\ \Leftrightarrow x\left(x-1\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
\(d,x^2\left(x-1\right)-4x^2+8x-4=0\\ \Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
a) \(x^4-2x^3+5x^2-10x=0\\ \Rightarrow\left(x^4-2x^3\right)+\left(5x^2-10x\right)=0\\ \Rightarrow x^3\left(x-2\right)+5x\left(x-2\right)=0\\ \Rightarrow\left(x^3+5x\right)\left(x-2\right)=0\\ \Rightarrow x\left(x^2+5\right)\left(x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x^2+5=0\\x-2=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=\pm\sqrt{5}\\x=2\end{matrix}\right.\)
Vậy \(x=\left\{-\sqrt{5};0;\sqrt{5};2\right\}\)
b) \(\left(3x+5\right)^2=\left(2x-2\right)^2\\ \Rightarrow\left[{}\begin{matrix}3x+5=2x-2\\3x+5=-2x+2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-7\\x=-\dfrac{3}{5}\end{matrix}\right.\)
c) \(x^3-2x^2+x=0\\ \Rightarrow x\left(x^2-2x+1\right)=0\\ \Rightarrow x\left(x-1\right)^2=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\\left(x-1\right)^2=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
vậy ...
d) \(x^2\left(x-1\right)-4x^2+8x-4=0\\ x^2\left(x-1\right)-\left(4x^2-8x+4\right)=0\\ x^2\left(x-1\right)-\left(2x-2\right)^2=0\\ \Rightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Rightarrow\left(x-1\right)\left[x^2-4\left(x-1\right)\right]=0\\ \Rightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Rightarrow\left(x-1\right)\left(x-2\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\\left(x-2\right)^2=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
a: Ta có: \(x^4-2x^3+5x^2-10x=0\)
\(\Leftrightarrow x\left(x^3-2x^2+5x-10\right)=0\)
\(\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
b:Ta có: \(\left(3x+5\right)^2=\left(2x-2\right)^2\)
\(\Leftrightarrow\left(3x+5\right)^2-\left(2x-2\right)^2=0\)
\(\Leftrightarrow\left(3x+5-2x+2\right)\left(3x+5+2x-2\right)=0\)
\(\Leftrightarrow\left(x+7\right)\left(5x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-7\\x=-\dfrac{3}{5}\end{matrix}\right.\)
tìm x:
x^3-16x=0
x^4-2x^3+10x^2-20x=0
(2x-3)^2=(x+5)^2
x^2(x-1)-4x^2+8x-4=0
*\(\left(2x-3\right)^2=\left(x+5\right)^2\)
\(\Rightarrow\left(2x-3\right)^2-\left(x+5\right)^2=0\)
\(\Rightarrow\left(2x-3-x-5\right)\left(2x-3+x+5\right)=0\)
\(\Rightarrow\left(x-8\right)\left(3x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=8\\x=-\dfrac{2}{3}\end{matrix}\right.\)
* \(x^3-16x=0\)
\(\Rightarrow x\left(x^2-16\right)=0\)
\(\Rightarrow x\left(x^2-4^2\right)=0\)
\(\Rightarrow x\left(x-4\right)\left(x+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
*\(x^4-2x^3+10x^2-20x=0\)
\(\Rightarrow\left(x^4+10x^2\right)-\left(2x^3+20x\right)=0\)
\(\Rightarrow x^2\left(x^2+10\right)-2x\left(x^2+10\right)=0\)
\(\Rightarrow\left(x^2+10\right)\left(x^2-2x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\varnothing\\x=2\end{matrix}\right.\)
I) THỰC HIỆN PHÉP TÍNH a) 2x(x^2-4y) b)3x^2(x+3y) c) -1/2x^2(x-3) d) (x+6)(2x-7)+x e) (x-5)(2x+3)+x II phân tích đa thức thành nhân tử a) 6x^2+3xy b) 8x^2-10xy c) 3x(x-1)-y(1-x) d) x^2-2xy+y^2-64 e) 2x^2+3x-5 f) 16x-5x^2-3 g) x^2-5x-6 IIITÌM X BIẾT a)2x+1=0 b) -3x-5=0 c) -6x+7=0 d)(x+6)(2x+1)=0 e)2x^2+7x+3=0 f) (2x-3)(2x+1)=0 g) 2x(x-5)-x(3+2x)=26 h) 5x(x-1)=x-1 IV TÌM GTNN,GTLN. a) tìm giá trị nhỏ nhất x^2-6x+10 2x^2-6x b) tìm giá trị lớn nhất 4x-x^2-5 4x-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
^^
Phân tích đa thức thành nhân tử
a) x³-3x²+3x-1-8y³
b) x⁴-4x³+8x²-16x+16
Giải pt
a) 6(x-3) +(x-1) ²-(x+1) ²=2x
b) (x+4) ²-(x+8) (x-8) =96
c) 4x²-1=(2x+1) (3x-5)
d) 2x²-x=3-6x
e) 2x³+5x²-3x=0
f) x(2x-7) -4x+14=0
g) (2x-5) ²-(x+2) ²=0
h) (3x+1) (7x+3) =(5x-7) (3x+1)
i) x²+10x+25-4x(x+5) =0
k))(4x-5) ²-2(16x²-25) =0
l) (4x+3) ²=4(x²-2x+1)
m) x²-11x+28=0
n) 3x³-3x²-6x=0
o) x²-9x+20=0
\(o,x^2-9x+20=0\)
\(\Leftrightarrow x^2-4x-5x+20=0\)
\(\Leftrightarrow x\left(x-4\right)-5\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x-5=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=4\\x=5\end{cases}}\)
\(n,3x^3-3x^2-6x=0\)
\(\Leftrightarrow3x\left(x^2-x-2\right)=0\)
\(\Leftrightarrow3x\left(x^2+x-2x-2\right)=0\)
\(\Leftrightarrow3x\left[x\left(x+1\right)-2\left(x+1\right)\right]=0\)
\(\Leftrightarrow3x\left(x+1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\orbr{\begin{cases}3x=0\\x+1=0\end{cases}}\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\orbr{\begin{cases}x=0\\x=-1\end{cases}}\\x=2\end{cases}}\)
\(m,x^2-11x+28=0\)
\(\Leftrightarrow x^2-4x-7x+28=0\)
\(\Leftrightarrow x\left(x-4\right)-7\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-7\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x-7=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=4\\x=7\end{cases}}\)
\(l,\left(4x+3\right)^2=4\left(x^2-2x+1\right)\)
\(\Leftrightarrow16x^2+24x+9=4x^2-8x+4\)
\(\Leftrightarrow16x^2+24x+9-4x^2+8x-4=0\)
\(\Leftrightarrow12x^2+32x+5=0\)
\(\Leftrightarrow\left(x+\frac{1}{6}\right)\left(x+\frac{5}{2}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{6}=0\\x+\frac{5}{2}=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-\frac{1}{6}\\x=-\frac{5}{2}\end{cases}}\)
tìm x
a)(x+6)^2-x(x+9)=0
b)6x(2x+5)-(3x+4)(4x-3)=9
c)2x(8x+3)-(4x+1)=13
d)(x-4)^2-x(x+4)=0
e)(x-2)^2-(2x+3)(x-2)=0
a) \(\left(x+6\right)^2-x\left(x+9\right)=0\)
\(\Leftrightarrow\)\(x^2+12x+36-x^2-9x=0\)
\(\Leftrightarrow\)\(3x+36=0\)
\(\Leftrightarrow\)\(x=-12\)
Vậy...
b) \(6x\left(2x+5\right)-\left(3x+4\right)\left(4x-3\right)=9\)
\(\Leftrightarrow\)\(12x^2+30x-12x^2-7x+12=9\)
\(\Leftrightarrow\)\(23x+12=9\)
\(\Leftrightarrow\)\(x=-\frac{3}{23}\)
Vậy
c) \(2x\left(8x+3\right)-\left(4x+1\right)=13\)
\(\Leftrightarrow\)\(16x^2+6x-4x-1=13\)
\(\Leftrightarrow\)\(16x^2+2x-14=0\)
\(\Leftrightarrow\)\(8x^2+x-7=0\)
\(\Leftrightarrow\)\(\left(x+1\right)\left(8x-7\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-1\\x=\frac{7}{8}\end{cases}}\)
Vậy
d) \(\left(x-4\right)^2-x\left(x+4\right)=0\)
\(\Leftrightarrow\)\(x^2-8x+16-x^2-4x=0\)
\(\Leftrightarrow\)\(-12x+16=0\)
\(\Leftrightarrow\)\(x=\frac{4}{3}\)
Vậy
e) \(\left(x-2\right)^2-\left(2x+3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\)\(x^2-4x+4-2x^2+x+6=0\)
\(\Leftrightarrow\)\(-x^2-3x+10=0\)
\(\Leftrightarrow\)\(\left(2-x\right)\left(x+5\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=2\\x=-5\end{cases}}\)
Vậy
giải pt
a, 2x^3++3x^2-8x-12=0
b, x^3-4x^2-x+4=0
c,x^3-x^2-x-2=0
d,x^4-3x^3+3x^2-x=0
e,(x+1)(x^2-2x+3)=x^3+1
g,x^3+3x^2+3x+1=4x+4
a) \(2x^3+3x^2-8x-12=0\)
\(\Leftrightarrow\left(2x^3-8x\right)+\left(3x^2-12\right)=0\)
\(\Leftrightarrow2x\left(x^2-4\right)+3\left(x^2-4\right)=0\)
\(\Leftrightarrow\left(x^2-4\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\)\(x-2=0\)
hoặc \(x+2=0\)
hoặc \(2x+3=0\)
\(\Leftrightarrow\)\(x=2\)
hoặc \(x=-2\)
hoặc \(x=-\frac{3}{2}\)
Vậy tập nghiệm của phương trình là \(S=\left\{2;-2;-\frac{3}{2}\right\}\)
b) \(x^3-4x^2-x+4=0\)
\(\Leftrightarrow x^2\left(x-4\right)-\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\)\(x-4=0\)
hoặc \(x-1=0\)
hoặc \(x+1=0\)
\(\Leftrightarrow\)\(x=4\)
hoặc \(x=1\)
hoặc \(x=-1\)
Vậy tập nghiệm của phương trình là \(S=\left\{4;1;-1\right\}\)
c) \(x^3-x^2-x-2=0\)
\(\Leftrightarrow x^3-2x^2+x^2-2x+x-2=0\)
\(\Leftrightarrow x^2\left(x-2\right)+x\left(x-2\right)+\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x^2+x+1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\left(tm\right)\\\left(x+\frac{1}{2}\right)^2+\frac{3}{4}=0\left(ktm\right)\end{cases}}\)
Vậy tập nghiệm của phương trình là \(S=\left\{2\right\}\)
d) \(x^4-3x^3+3x^2-x=0\)
\(\Leftrightarrow x\left(x^3-3x^2+3x-1\right)=0\)
\(\Leftrightarrow x\left(x-1\right)^3=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
Vậy tập nghiệm của phương trình là \(S=\left\{0;1\right\}\)
e) \(\left(x+1\right)\left(x^2-2x+3\right)=x^3+1\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-2x+3\right)=\left(x+1\right)\left(x^2-x+1\right)\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x^2-2x+3=x^2-x+1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=2\end{cases}}\)
Vậy tập nghiệm của phương trình là \(S=\left\{-1;2\right\}\)
g) \(x^3+3x^2+3x+1=4x+4\)
\(\Leftrightarrow\left(x+1\right)^3=4\left(x+1\right)\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\\left(x+1\right)^2=4\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x+1=\pm2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-3\end{cases}}\) hoặc \(x=1\)
Vậy tập nghiệm của phương trình là \(S=\left\{-1;1;-3\right\}\)
b) \(x^3-4x^2-x+4=0\)
\(\Leftrightarrow x^2\left(x-4\right)-\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x^2-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=4\\x=\pm1\end{cases}}\)
c) \(x^3-x^2-x-2=0\)
\(\Leftrightarrow x^3-2x^2+x^2-2x+x-2=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+x+1\right)=0\)
\(\Leftrightarrow x=2\) ( Do \(x^2+x+1>0\) )
a) \(2x^3+3x^2-8x-12=0\)
\(\Leftrightarrow x^2\left(2x+3\right)-4\left(2x+3\right)=0\)
\(\Leftrightarrow\left(2x+3\right)\left(x^2-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x+3=0\\x^2-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{3}{2}\\x=\pm2\end{cases}}\)
1)
a) 2-x/2001 - 1=1-x/2002 - x/2003
b)x^3 + 3x^2 + x + 3=0
c)/x-4/=/3-2x/
d)4X^2 + 16x +17
e)13-2/3x+2/=-1
f)/3x-4/=x-5
2)
a) tìm x thuộc Z để A=3x^2 - 9x + 2/x-3 thuộc Z
b)với giá trị nào của n thuộc Z thì A=3n+9/n-4 thuộc Z
3) chứng minh các bất phương trình sau vô nghiệm
a)x^2+x+2 nhỏ hơn bằng 0 b)x^4-2x^2+5 nhỏ hơn bằng 0
4)
1) x^4-8x^3+11x^2+8x-12=0 2)-3x^4+20x^3-35x^2-10x+48=0
3)x^5-5x^4+6x^3-x^2+5x-6=0 4)(x^2+x+1)(x^2+x+2)=12
5)(x-3)(x-5)(x-6)(x-10)-24x^2=0 6)(x+1)(x-4)(x+2)(x-8)+4x^2=0
7)(x^2-4)(x^2-10)=72
tìm x
a)(x+6)^2-x(x+9)=0
b)6x(2x+5)-(3x+4)(4x-3)=9
c)2x(8x+3)-(4x+1)=13
d)(x-4)^2-x(x+4)=0
e)(x-2)^2-(2x+3)(x-2)=0a)x^3+4x+x-6=0
b)2x^3+3x^2-9x-10=0
c)x^3-x-6=0
d)(x^2-4x)^2+(8x-4)=-15
e)x(x+1)(x-1)(x+2)=24
g)(x^2-3x+3)(x^2-2x+3)=2x^2
h)(x+1)^4 x (x+3)^4=272
a)x^3+5x-6=0
x^3-x+6x-6
x(x^2-1)+6(x-1)=0
(x-1)(x(x+1)+6)
(x-1)(x^2+x+6)=0
=>x-1 hoac x^2+x+6
TH1
x-1=0
x=1
TH2
x^2+x+6=0
(x^2+2.1/2 x +1/4)-1/4+6=0
(x+1/2)^2=-23/4
vi (x+1/2)^2 >/0
=>pt vo nghiem
=>nghiem cua pt la 1