Tìm x biết
x^3 + 6x^2 +12x= 19
5(x + 9)^2(x - 4)^3 - 10(x + 9)^3(x - 4)^2 = 0
(2x + 3)^2 + (x - 2)^2 - 2(2x +3 )(x - 2)
6) \(\sqrt{x^2+12x+36}=-x-6\)
7) \(\sqrt{9x^2-12x+4}=3x-2\)
8) \(\sqrt{16-24x+9x^2}=2x-10\)
9) \(\sqrt{x^2-6x+9}==2x-3\)
10) \(\sqrt{x^2-3x+\dfrac{9}{4}}=\dfrac{3}{x}x-4\)
6) ĐKXĐ: \(x\le-6\)
\(\sqrt{\left(x+6\right)^2}=-x-6\Leftrightarrow\left|x+6\right|=-x-6\)
\(\Leftrightarrow x+6=x+6\left(đúng\forall x\right)\)
Vậy \(x\le-6\)
7) ĐKXĐ: \(x\ge\dfrac{2}{3}\)
\(pt\Leftrightarrow\sqrt{\left(3x-2\right)^2}=3x-2\Leftrightarrow\left|3x-2\right|=3x-2\)
\(\Leftrightarrow3x-2=3x-2\left(đúng\forall x\right)\)
Vậy \(x\ge\dfrac{2}{3}\)
8) ĐKXĐ: \(x\ge5\)
\(pt\Leftrightarrow\sqrt{\left(4-3x\right)^2}=2x-10\)\(\Leftrightarrow\left|4-3x\right|=2x-10\)
\(\Leftrightarrow4-3x=10-2x\Leftrightarrow x=-6\left(ktm\right)\Leftrightarrow S=\varnothing\)
9) ĐKXĐ: \(x\ge\dfrac{3}{2}\)
\(pt\Leftrightarrow\sqrt{\left(x-3\right)^2}=2x-3\Leftrightarrow\left|x-3\right|=2x-3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=2x-3\left(x\ge3\right)\\x-3=3-2x\left(\dfrac{3}{2}\le x< 3\right)\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=2\left(tm\right)\end{matrix}\right.\)
Bài 3 : Tìm x biết
a) (x-2)^2-x(x-3)=0
b) (x+3)(2x+1)-2(x-1)^2=0
c) (4x-5)^2=9(2-5x)^2
d) X^2-6x-13=0
e) (x+2)(x^2-2x+4)-x(x^2+2)=15
f) X^3-6x^2+12x-19=0
e: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)
\(\Leftrightarrow x^3+8-x^3-2x=15\)
\(\Leftrightarrow2x=-7\)
hay \(x=-\dfrac{7}{2}\)
f: Ta có: \(x^3-6x^2+12x-19=0\)
\(\Leftrightarrow x^3-6x^2+12x-8-11=0\)
\(\Leftrightarrow\left(x-2\right)^3=11\)
hay \(x=\sqrt[3]{11}+2\)
Giải phương trình:
1. \(x^4-6x^2-12x-8=0\)
2. \(\dfrac{x}{2x^2+4x+1}+\dfrac{x}{2x^2-4x+1}=\dfrac{3}{5}\)
3. \(x^4-x^3-8x^2+9x-9+\left(x^2-x+1\right)\sqrt{x+9}=0\)
4. \(2x^2.\sqrt{-4x^4+4x^2+3}=4x^4+1\)
5. \(x^2+4x+3=\sqrt{\dfrac{x}{8}+\dfrac{1}{2}}\)
6. \(\left\{{}\begin{matrix}4x^3+xy^2=3x-y\\4xy+y^2=2\end{matrix}\right.\)
7. \(\left\{{}\begin{matrix}\sqrt{x^2-3y}\left(2x+y+1\right)+2x+y-5=0\\5x^2+y^2+4xy-3y-5=0\end{matrix}\right.\)
8. \(\left\{{}\begin{matrix}\sqrt{2x^2+2}+\left(x^2+1\right)^2+2y-10=0\\\left(x^2+1\right)^2+x^2y\left(y-4\right)=0\end{matrix}\right.\)
1.
\(x^4-6x^2-12x-8=0\)
\(\Leftrightarrow x^4-2x^2+1-4x^2-12x-9=0\)
\(\Leftrightarrow\left(x^2-1\right)^2=\left(2x+3\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-1=2x+3\\x^2-1=-2x-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-4=0\\x^2+2x+2=0\end{matrix}\right.\)
\(\Leftrightarrow x=1\pm\sqrt{5}\)
3.
ĐK: \(x\ge-9\)
\(x^4-x^3-8x^2+9x-9+\left(x^2-x+1\right)\sqrt{x+9}=0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left(\sqrt{x+9}+x^2-9\right)=0\)
\(\Leftrightarrow\sqrt{x+9}+x^2-9=0\left(1\right)\)
Đặt \(\sqrt{x+9}=t\left(t\ge0\right)\Rightarrow9=t^2-x\)
\(\left(1\right)\Leftrightarrow t+x^2+x-t^2=0\)
\(\Leftrightarrow\left(x+t\right)\left(x-t+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-t\\x=t-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{x+9}\\x=\sqrt{x+9}-1\end{matrix}\right.\)
\(\Leftrightarrow...\)
2.
ĐK: \(x\ne\dfrac{2\pm\sqrt{2}}{2};x\ne\dfrac{-2\pm\sqrt{2}}{2}\)
\(\dfrac{x}{2x^2+4x+1}+\dfrac{x}{2x^2-4x+1}=\dfrac{3}{5}\)
\(\Leftrightarrow\dfrac{1}{2x+\dfrac{1}{x}+4}+\dfrac{1}{2x+\dfrac{1}{x}-4}=\dfrac{3}{5}\)
Đặt \(2x+\dfrac{1}{x}+4=a;2x+\dfrac{1}{x}-4=b\left(a,b\ne0\right)\)
\(pt\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{3}{5}\left(1\right)\)
Lại có \(a-b=8\Rightarrow a=b+8\), khi đó:
\(\left(1\right)\Leftrightarrow\dfrac{1}{b+8}+\dfrac{1}{b}=\dfrac{3}{5}\)
\(\Leftrightarrow\dfrac{2b+8}{\left(b+8\right)b}=\dfrac{3}{5}\)
\(\Leftrightarrow10b+40=3\left(b+8\right)b\)
\(\Leftrightarrow\left[{}\begin{matrix}b=2\\b=-\dfrac{20}{3}\end{matrix}\right.\)
TH1: \(b=2\Leftrightarrow...\)
TH2: \(b=-\dfrac{20}{3}\Leftrightarrow...\)
Câu 1: giải các phương trình ;
a, x.(2x-9)=3x.(x-5) b,(2x-1)^2-(2x+1)^2=4.(x-3) c,2x+3/3+3x+2/2=2,5x-1 d,2-x/2001-1=1-x/2002-x/2003
e, x+1/3+3.(2x+1)/4=2x+3.(x+1)/6+7+12x/12
Câu 2:giải các phương trình sau:
a,(x^2-6x+9)^2-15(x^2-6x+10)=1 b,(x^2+1)+3x(x^2+1)+2x^2=0 c,(x^2-9)^2=12x+1 d,x63=3x62=4x=2=0 e,x^4+x^2+6x-8=0 g, (x^2-4x)2+(x-2)62=10 h,(12x+7)(3x+2)(2x+1)=3 i,(x^2+5x+4)(9x^2+30x+16)=4x^2 k, (x^2+x+1)^2=3(x^4+x^2+1) l, 6x^4+25x^3+12x^2-25x+6=0
Tìm x,y,z
a) (x-3)^3-(x-3)(x^2+3x+9)+9(x+1)^2=15
b) (x^2-2)^2+4(x-1)^2-4(x^2-2)(x-1)=0
c) x^2+y^2+z^2= 4x-2y+6z-14
d) 8x^3-12x^2+6x-1=0
e) x^2+5y^2-4xy-8y+2x+5=0
f) x(x-5)(x+5)-(x-2)(x^2+2x+4)=3
Tìm x biết
a)(x+3)^2(x-2)^2=2x b)7x(x-2)=(x-2) c)8x^3-12x^2+6x-1=0
d)4x^2-9-x(2x-3)=0 e)x^3+5x^2+9x=-45 f)x^3-6x^2-x+30=0
d) \(4x^2-9-x\left(2x-3\right)=0\)
\(\Leftrightarrow4x^2-9-2x^2+3x=0\)
\(\Leftrightarrow2x^2+3x-9=0\)
\(\Delta=3^2-4.2.\left(-9\right)=9+72=81\)
Vậy pt có 2 nghiệm phân biệt
\(x_1=\frac{-3+\sqrt{81}}{4}=\frac{-3}{2}\);\(x_1=\frac{-3-\sqrt{81}}{4}=-3\)
e) \(x^3+5x^2+9x=-45\)
\(\Leftrightarrow x^3+5x^2+9x+45=0\)
\(\Leftrightarrow x^2\left(x+5\right)+9\left(x+5\right)=0\)
\(\Leftrightarrow\left(x^2+9\right)\left(x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2+9=0\\x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\pm3i\\x=-5\end{cases}}\)
f) \(x^3-6x^2-x+30=0\)
\(\Leftrightarrow\left(x^3-x^2-6x\right)-\left(5x^2-5x-30\right)=0\)
\(\Leftrightarrow x\left(x^2-x-6\right)-5\left(x^2-x-6\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x^2-x-6\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x^2-2x+3x-6\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left[x\left(x-2\right)+3\left(x-2\right)\right]=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+3\right)\left(x-2\right)=0\)
\(\Leftrightarrow x\in\left\{5;-3;2\right\}\)
Bài 5: Giải phương trình
a) x ( x+1) ( x – 1) ( x+2) =24
b) ( x – 4) ( x – 5) ( x – 6) ( x – 7) = 1680
c) 2x ( 8x – 1) 2 ( 4x – 1) = 9
d) ( 2x + 1)( x + 1) 2 ( 2x + 3) = 18
e) (x 2 – 6x +9 ) 2 – 15( x 2 – 6x +10) = 1
f) ( x 2 + 1) 2 + 3x ( x 2 + 1) + 2x 2 = 0
g) ( x 2 – 9) 2 = 12x + 1
Bài 6: Giải phương trình
a) ( x + 3) 4 + ( x+ 5 ) 4 = 16
b) ( x – 2) 4 + ( x – 3) 4 = 1
Bài 5: Giải phương trình
a) x ( x+1) ( x – 1) ( x+2) =24
b) ( x – 4) ( x – 5) ( x – 6) ( x – 7) = 1680
c) 2x ( 8x – 1) 2 ( 4x – 1) = 9
d) ( 2x + 1)( x + 1) 2 ( 2x + 3) = 18
e) (x 2 – 6x +9 ) 2 – 15( x 2 – 6x +10) = 1
f) ( x 2 + 1) 2 + 3x ( x 2 + 1) + 2x 2 = 0
g) ( x 2 – 9) 2 = 12x + 1
Bài 6: Giải phương trình
a) ( x + 3) 4 + ( x+ 5 ) 4 = 16
b) ( x – 2) 4 + ( x – 3) 4 = 1
1. 6x(x - 10) - 2x+20=0 6. 3x2 - 6x+3=0
2. 3x2(x - 3) + 3(3 - x)=0 7. 4x2 - 10x+2=0
3. x2 - 8x+16=2(x -4) 8. x2 - 12x -18=0
4. x2 - 16 + 7x ( x+4)=0 9. 3x2 - 10x+3=0
5. x2 - 13x - 14=0 10. 5x2 - 10x+10=0
\(1.6x\left(x-10\right)-2x+20=0\)
⇔\(6x\left(x-10\right)-2\left(x-10\right)=0\)
⇔ \(2\left(x-10\right)\left(3x-1\right)=0\)
⇔ x = 10 hoặc x = \(\dfrac{1}{3}\)
KL....
\(2.3x^2\left(x-3\right)+3\left(3-x\right)=0\)
⇔ \(3\left(x-3\right)\left(x^2-1\right)=0\)
⇔ \(x=+-1\) hoặc \(x=3\)
KL....
\(3.x^2-8x+16=2\left(x-4\right)\)
⇔ \(\left(x-4\right)^2-2\left(x-4\right)=0\)
⇔ \(\left(x-4\right)\left(x-6\right)=0\)
⇔ \(x=4\) hoặc \(x=6\)
KL.....
\(4.x^2-16+7x\left(x+4\right)=0\)
\(\text{⇔}4\left(x+4\right)\left(2x-1\right)=0\)
⇔ \(x=-4hoacx=\dfrac{1}{2}\)
KL.....
\(5.x^2-13x-14=0\)
⇔ \(x^2+x-14x-14=0\)
\(\text{⇔}\left(x+1\right)\left(x-14\right)=0\)
\(\text{⇔}x=14hoacx=-1\)
KL......
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