\(e.\left(x-2\right).\left(x+2\right)<0\)\(d.\left(_{-x^2-7}\right).\left(x+1\right)>0\)\(c.\left(x^2+5\right).\left(x-3\right)>0\)\(a.\left(x^2-3\right).\left(2x+10\right)=0\)\(b.\left(x-2\right)^2-25=0\)
Tìm x, biết :
a/ \(\dfrac{1}{3}x\left(x^2-4\right)=0\)
b/ \(x\left(x+5\right)=x+5\)
c/ \(x^3-\dfrac{1}{9}x=0\)
3)\(^2-\left(x+5\right)^2=0\)
e/ \(\left(x+2\right)^2-\left(x-2\right)\left(x+2\right)=0\)
f/ \(x\left(2x-3\right)-6+4x=0\)
g/ \(2\left(3x-2\right)^2-9x^2+4=0\)
h/ \(x^2\left(x+1\right)+2x\left(x+1\right)=0\)
i/ \(4x^2+9x+5=0\)
a) \(\Rightarrow\dfrac{1}{3}x\left(x-2\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
b) \(\Rightarrow\left(x+5\right)\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=-5\\x=1\end{matrix}\right.\)
c) \(\Rightarrow x\left(x^2-\dfrac{1}{9}\right)=0\Rightarrow x\left(x-\dfrac{1}{3}\right)\left(x+\dfrac{1}{3}\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)
e) \(\Rightarrow\left(x+2\right)\left(x+2-x+2\right)=0\Rightarrow\left(x+2\right).4=0\Rightarrow x=-2\)
f) \(\Rightarrow x\left(2x-3\right)+2\left(2x-3\right)=0\Rightarrow\left(2x-3\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-2\end{matrix}\right.\)
g) \(\Rightarrow2\left(3x-2\right)^2-\left(3x-2\right)\left(3x+2\right)=0\Rightarrow\left(3x-2\right)\left(3x-6\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=2\end{matrix}\right.\)
h) \(\Rightarrow x\left(x+1\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=-2\end{matrix}\right.\)
i) \(\Rightarrow4x\left(x+1\right)+5\left(x+1\right)=0\Rightarrow\left(x+1\right)\left(4x+5\right)=0\Rightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{5}{4}\end{matrix}\right.\)
Bài Tập: Giải phương trình :
a) (x + 5)(2x - 3) = 0
b) \(\left(x^2-9\right)\left(4-x\right)=0\)
c) \(\left(2x+3\right)\left(4-5x\right)=0\)
d) \(2\left(x+3\right)\left(x-4\right)=0\)
e) \(\left(x^2-9\right)\left(4-x\right)=0\)
f) \(\left(2x+3\right)\left(x^2-16\right)=0\)
a: \(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=\dfrac{3}{2}\end{matrix}\right.\)
b: \(\Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\\x=4\end{matrix}\right.\)
c: \(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\5x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{4}{5}\end{matrix}\right.\)
d: \(\Leftrightarrow\left(x+3\right)\left(x-4\right)=0\)
=>x+3=0 hoặc x-4=0
=>x=-3 hoặc x=4
e: \(\Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\\x=4\end{matrix}\right.\)
f: \(\Leftrightarrow\left(2x+3\right)\left(x-4\right)\left(x+4\right)=0\)
hay \(x\in\left\{-\dfrac{3}{2};4;-4\right\}\)
a, \(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=\dfrac{3}{2}\end{matrix}\right.\)
b, \(\Leftrightarrow\left[{}\begin{matrix}x^2-9=0\\4-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\pm3\\x=4\end{matrix}\right.\)
c, \(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\4-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{4}{5}\end{matrix}\right.\)
d, \(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=4\end{matrix}\right.\)
e, tương tự d
f, \(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\x^2-16=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\pm4\end{matrix}\right.\)
Tìm \(x\):
\(8\)) \(1-\left(x-6\right)=4\left(2-2x\right)\)
\(9\))\(\left(3x-2\right)\left(x+5\right)=0\)
\(10\))\(\left(x+3\right)\left(x^2+2\right)=0\)
\(11\))\(\left(5x-1\right)\left(x^2-9\right)=0\)
\(12\))\(x\left(x-3\right)+3\left(x-3\right)=0\)
\(13\))\(x\left(x-5\right)-4x+20=0\)
\(14\))\(x^2+4x-5=0\)
\(8,1-\left(x-6\right)=4\left(2-2x\right)\)
\(\Leftrightarrow1-x+6=8-8x\)
\(\Leftrightarrow-x+8x=8-1-6\)
\(\Leftrightarrow7x=1\)
\(\Leftrightarrow x=\dfrac{1}{7}\)
\(9,\left(3x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-5\end{matrix}\right.\)
\(10,\left(x+3\right)\left(x^2+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x^2+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\varnothing\end{matrix}\right.\)
`8)1-(x-5)=4(2-2x)`
`<=>1-x+5=8-6x`
`<=>5x=2<=>x=2/5`
`9)(3x-2)(x+5)=0`
`<=>[(x=2/3),(x=-5):}`
`10)(x+3)(x^2+2)=0`
Mà `x^2+2 > 0 AA x`
`=>x+3=0`
`<=>x=-3`
`11)(5x-1)(x^2-9)=0`
`<=>(5x-1)(x-3)(x+3)=0`
`<=>[(x=1/5),(x=3),(x=-3):}`
`12)x(x-3)+3(x-3)=0`
`<=>(x-3)(x+3)=0`
`<=>[(x=3),(x=-3):}`
`13)x(x-5)-4x+20=0`
`<=>x(x-5)-4(x-5)=0`
`<=>(x-5)(x-4)=0`
`<=>[(x=5),(x=4):}`
`14)x^2+4x-5=0`
`<=>x^2+5x-x-5=0`
`<=>(x+5)(x-1)=0`
`<=>[(x=-5),(x=1):}`
\(11,=>\left[{}\begin{matrix}5x-1=0\\x^2-9=0\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=3\\x=-3\end{matrix}\right.\\ 12,=>\left(x+3\right)\left(x-3\right)=0\\ =>\left[{}\begin{matrix}x+3=0\\x-3=0\end{matrix}\right.=>\left[{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\\ 13,=>x\left(x-5\right)-4\left(x-5\right)=0\\ =>\left(x-4\right)\left(x-5\right)=0\\ =>\left[{}\begin{matrix}x-4=0\\x-5=0\end{matrix}\right.=>\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)
\(14,=>x^2+5x-x-5=0\\ =>x\left(x+5\right)-\left(x+5\right)=0\\ =>\left(x-1\right)\left(x+5\right)=0\\ =>\left[{}\begin{matrix}x-1=0\\x+5=0\end{matrix}\right.=>\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)
Tìm x:
a) \(3x\left(3x-8\right)-9x^2+8=0\)
b)\(6x-15-x\left(5-2x\right)=0\)
c) \(x^3-16x=0\)
d) \(2x^2+3x-5=0\)
e) \(3x^2-x\left(3x-6\right)=36\)
f) \(\left(x+2\right)^2-\left(x-5\right)\left(x+1\right)=17\)
g) \(\left(x-4\right)^2-x\left(x+6\right)=9\)
h) \(4x\left(x-1000\right)-x+1000=0\)
i) \(x^2-36=0\)
j) \(x^2y-2+x+x^2-2y+xy=0\)
k) \(x\left(x+1\right)-\left(x-1\right).\left(2x-3\right)=0\)
l) \(3x^3-27x=0\)
Giải các PT sau (Đặt ẩn phụ)
a,\(\left(x^2+x\right)^2+4\left(x^2+x\right)-12=0\)
b,\(\left(x^2+2x+3\right)^2-9\left(x^2+2x+3\right)+18=0\)
c,\(\left(x-2\right)\left(x+2\right)\left(x^2-10\right)=72\)
d,\(x\left(x+1\right)\left(x^2+x+1\right)=42\)
e,\(\left(x-1\right)\left(x-3\right)\left(x+5\right)\left(x+7\right)-297=0\)
f,\(x^4-2x^2-144x-1295=0\)
a, - Đặt \(x^2+x=a\) ta được phương trình :\(a^2+4a-12=0\)
=> \(a^2-2a+6a-12=0\)
=> \(a\left(a-2\right)+6\left(a-2\right)=0\)
=> \(\left(a+6\right)\left(a-2\right)=0\)
=> \(\left[{}\begin{matrix}a+6=0\\a-2=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}a=2\\a=-6\end{matrix}\right.\)
- Thay lại \(x^2+x=a\) vào phương trình trên ta được :\(\left[{}\begin{matrix}x^2+x=2\\x^2+x=-6\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x^2+x-2=0\\x^2+x+6=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}\left(x+\frac{1}{2}\right)^2-\frac{9}{4}=0\\\left(x+\frac{1}{2}\right)^2+\frac{23}{4}=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}\left(x+\frac{1}{2}\right)^2=\frac{9}{4}\\\left(x+\frac{1}{2}\right)^2=-\frac{23}{4}\left(VL\right)\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x+\frac{1}{2}=\sqrt{\frac{9}{4}}\\x+\frac{1}{2}=-\sqrt{\frac{9}{4}}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\sqrt{\frac{9}{4}}-\frac{1}{2}=1\\x=-\sqrt{\frac{9}{4}}-\frac{1}{2}=-2\end{matrix}\right.\)
Vậy phương trình trên có nghiệm là \(S=\left\{1,-2\right\}\)
b, Đặt \(x^2+2x+3=a\) -> làm tương tự câu a .
c, Ta có : \(\left(x-2\right)\left(x+2\right)\left(x^2-10\right)=72\)
=> \(\left(x^2-4\right)\left(x^2-10\right)=72\)
- Đặt \(x^2-4=a\) và \(x^2-10=a-6\) ta được phương trình :
\(a\left(a-6\right)=72\)
=> \(a^2-6a-72=0\)
=> \(a^2+6a-12a-72=0\)
=> \(a\left(a+6\right)-12\left(a+6\right)=0\)
=> \(\left(a+6\right)\left(a-12\right)=0\)
=> \(\left[{}\begin{matrix}a+6=0\\a-12=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}a=-6\\a=12\end{matrix}\right.\)
- Thay lại \(x^2-4=a\) vào phương trình trên ta được :\(\left[{}\begin{matrix}x^2-4=-6\\x^2-4=12\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x^2=-2\left(VL\right)\\x^2=16\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\sqrt{16}=4\\x=-\sqrt{16}=-4\end{matrix}\right.\)
Vậy phương trình trên có nghiệm là \(S=\left\{4,-4\right\}\)
d, Ta có : \(x\left(x+1\right)\left(x^2+x+1\right)=42\)
=> \(\left(x^2+x\right)\left(x^2+x+1\right)=42\)
- Đặt \(x^2+x=a\) ta được phương trình : \(a\left(a+1\right)=42\)
=> \(a^2+a-42=0\)
=> \(a^2+7a-6a-42=0\)
=> \(a\left(a+7\right)-6\left(a+7\right)=0\)
=> \(\left(a-6\right)\left(a+7\right)=0\)
=> \(\left[{}\begin{matrix}a=6\\a=-7\end{matrix}\right.\)
- Thay \(a=x^2+x\) vào phương trình ta được : \(\left[{}\begin{matrix}x^2+x=6\\x^2+x=-7\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x^2+x-6=0\\x^2+x+7=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}\left(x+\frac{1}{2}\right)^2-\frac{25}{4}=0\\\left(x+\frac{1}{2}\right)^2+\frac{27}{4}=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}\left(x+\frac{1}{2}\right)^2=\frac{25}{4}\\\left(x+\frac{1}{2}\right)^2=-\frac{27}{4}\left(VL\right)\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x+\frac{1}{2}=\sqrt{\frac{25}{4}}\\x+\frac{1}{2}=-\sqrt{\frac{25}{4}}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\sqrt{\frac{25}{4}}-\frac{1}{2}=2\\x=-\sqrt{\frac{25}{4}}-\frac{1}{2}=-3\end{matrix}\right.\)
Vậy phương trình trên có tập nghiệm là \(S=\left\{2;-3\right\}\)
A)\(\left(2x-7\right)^2-6\left(2x-7\right)\left(x-3\right)=0\)
B) \(x^3+1+\left(x^2-x+1\right)=0\)
C) \(\left(3x+1\right)^2-x^2+8x-16=0\)
D)\(\left(x+1\right)\left(x-1\right)^2\left(x+1\right)\left(x-2\right)^2=0\)
E) \(\left(3x+1\right)\left(x-3\right)^2=\left(3x+1\right)\left(2x-5\right)^2\)
F) \(\left(x+5\right)\left(3x+2\right)^2=x^2\left(x+5\right)\)
Tìm x , biết :
a. \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=15\)
b. \(2x^3-50x=0\)
c.\(5x^2-4\left(x^2-2x+1\right)-5=0\)
d. \(x^3-x=0\)
e. \(27x^3-27x^2+9x-1=1\)
a) Ta có: \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=15\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6\left(x^2+2x+1\right)=15\)
\(\Leftrightarrow-6x^2+12x+19+6x^2+12x+6=15\)
\(\Leftrightarrow24x+25=15\)
\(\Leftrightarrow24x=-10\)
hay \(x=-\dfrac{5}{12}\)
b) Ta có: \(2x^3-50x=0\)
\(\Leftrightarrow2x\left(x-5\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)
c) Ta có: \(5x^2-4\left(x^2-2x+1\right)-5=0\)
\(\Leftrightarrow5x^2-4x^2+8x-4-5=0\)
\(\Leftrightarrow x^2+8x-9=0\)
\(\Leftrightarrow\left(x+9\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=1\end{matrix}\right.\)
d) Ta có: \(x^3-x=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
e) Ta có: \(27x^3-27x^2+9x-1=1\)
\(\Leftrightarrow\left(3x\right)^3-3\cdot\left(3x\right)^2\cdot1+3\cdot3x\cdot1^2-1^3=1\)
\(\Leftrightarrow\left(3x-1\right)^3=1\)
\(\Leftrightarrow3x-1=1\)
\(\Leftrightarrow3x=2\)
hay \(x=\dfrac{2}{3}\)
Tìm x biết :
a) \(\left(x-2\right)^3+6\left(x+1\right)^2-x^3+12=0\)
b) \(\left(x-5\right)\left(x+5\right)-\left(x+3\right)^3+3\left(x-2\right)^2=\left(x+1\right)^2-\left(x+4\right)\left(x-4\right)+3x^2\)
c) \(\left(2x+3\right)^2+\left(x-1\right)\left(x+1\right)=5\left(x+2\right)^2-\left(x-5\right)\left(x+1\right)+\left(x+4\right)^2\)
d) \(\left(1-3x\right)^2-\left(x-2\right)\left(9x+1\right)=\left(3x-4\right)\left(3x+4\right)-9\left(x+3\right)^2\)
a/ \(x=\dfrac{-5}{12}\)
b/ \(x\approx-1,9526\)
c/ \(x=\dfrac{21-i\sqrt{199}}{10}\)
d/ \(x=\dfrac{-20}{13}\)
a) (x-2)3+6(x+1)2-x3+12=0
⇒ x3-6x2+12x-8+6(x2+2x+1)-x3+12=0
⇒ x3-6x2+12x-8+6x2+12x+6-x3+12=0
⇒ 24x+10=0
⇒ 24x=-10
⇒ x=-5/12
a.
PT \(\Leftrightarrow x^3-6x^2+12x-8+6(x^2+2x+1)-x^3+12=0\)
\(\Leftrightarrow x^3-6x^2+12x-8+6x^2+12x+6-x^3+12=0\)
\(\Leftrightarrow 24x+10=0\Leftrightarrow x=\frac{-5}{12}\)
b. Bạn xem lại đề, nghiệm khá xấu không phù hợp với mức độ tổng thể của bài.
c.
PT $\Leftrightarrow (4x^2+12x+9)+(x^2-1)=5(x^2+4x+4)+(x^2-4x-5)+9(x^2+6x+9)$
$\Leftrightarrow 10x^2+42x+64=0$
$\Leftrightarrow x^2+(3x+7)^2=-15< 0$ (vô lý)
Do đó pt vô nghiệm.
d.
PT $\Leftrightarrow (1-6x+9x^2)-(9x^2-17x-2)=(9x^2-16)-9(x^2+6x+9)$
$\Leftrightarrow 11x+3=-54x-97$
$\Leftrightarrow 65x=-100$
$\Leftrightarrow x=\frac{-20}{13}$
i, \(\left(x-1\right)\left(x+3\right)-\left(x-1\right)\left(2x+1\right)=0\)
k, \(\left(x+2\right)\left(x+1\right)-\left(x-3\right)\left(x+2\right)=0\)
l, \(\left(x-2\right)\left(x+3\right)=\left(x-2\right)\left(2x+5\right)\)
\(\left(x-1\right)\left(-x+2\right)=0\Leftrightarrow x=1;x=2\)
\(\left(x+2\right)\left(x+1-x+3\right)=0\Leftrightarrow x=-2\)
\(\left(x-2\right)\left(x+3\right)-\left(x-2\right)\left(2x+5\right)=0\Leftrightarrow\left(x-2\right)\left(-x-2\right)=0\Leftrightarrow x=-2;x=2\)
\(i,\left(x-1\right)\left(x+3\right)-\left(x-1\right)\left(2x+1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+3-2x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(-x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\\ k,\left(x+2\right)\left(x+1\right)-\left(x-3\right)\left(x+2\right)=0\\ \Leftrightarrow\left(x+2\right)\left(x+1-x+3\right)=0\\ \Leftrightarrow4\left(x+2\right)=0\\ \Leftrightarrow x+2=0\\ \Leftrightarrow x=-2\\ l,\left(x-2\right)\left(x+3\right)=\left(x-2\right)\left(2x+5\right)\\ \Leftrightarrow\left(x-2\right)\left(2x+5\right)-\left(x-2\right)\left(x+3\right)=0\\ \Leftrightarrow\left(x-2\right)\left(2x+5-x-3\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)