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chẳng ai giải, thôi mình giải vậy!
a) Đặt \(y=x^2+4x+8\),phương trình có dạng:
\(t^2+3x\cdot t+2x^2=0\)
\(\Leftrightarrow t^2+xt+2xt+2x^2=0\)
\(\Leftrightarrow t\left(t+x\right)+2x\left(t+x\right)=0\)
\(\Leftrightarrow\left(2x+t\right)\left(t+x\right)=0\)
\(\Leftrightarrow\left(2x+x^2+4x+8\right)\left(x^2+4x+8+x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-4\end{cases}}\)vậy tập nghiệm của phương trình là:S={-2;-4}
b) nhân 2 vế của phương trình với 12 ta được:
\(\left(6x+7\right)^2\left(6x+8\right)\left(6x+6\right)=72\)
Đặt y=6x+7, ta được:\(y^2\left(y+1\right)\left(y-1\right)=72\)
giải tiếp ra ta sẽ được S={-2/3;-5/3}
c) \(\left(x-2\right)^4+\left(x-6\right)^4=82\)
S={3;5}
d)s={1}
e) S={1;-2;-1/2}
f) phương trình vô nghiệm
a: \(\Leftrightarrow x^2\left(x^2+x-12\right)=0\)
\(\Leftrightarrow x^2\left(x+4\right)\left(x-3\right)=0\)
hay \(x\in\left\{0;-4;3\right\}\)
d: \(\left(x^2+5x\right)^2-2\left(x^2+5x\right)-24=0\)
\(\Leftrightarrow\left(x^2+5x-6\right)\left(x^2+5x+4\right)=0\)
\(\Leftrightarrow\left(x+6\right)\left(x-1\right)\left(x+1\right)\left(x+4\right)=0\)
hay \(x\in\left\{-6;1;-1;-4\right\}\)
f: \(x\left(x+1\right)\left(x-1\right)\left(x+2\right)=24\)
\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)=24\)
\(\Leftrightarrow\left(x^2+x\right)^2-2\left(x^2+x\right)-24=0\)
\(\Leftrightarrow x^2+x-6=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-2\right)=0\)
hay \(x\in\left\{-3;2\right\}\)
a) Ta có: \(2x^3+5x^2-3x=0\)
\(\Leftrightarrow x\left(2x^2+5x-3\right)=0\)
\(\Leftrightarrow x\left(2x^2+6x-x-3\right)=0\)
\(\Leftrightarrow x\left[2x\left(x+3\right)-\left(x+3\right)\right]=0\)
\(\Leftrightarrow x\left(x+3\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{0;-3;\dfrac{1}{2}\right\}\)
b) Ta có: \(2x^3+6x^2=x^2+3x\)
\(\Leftrightarrow2x^2\left(x+3\right)=x\left(x+3\right)\)
\(\Leftrightarrow2x^2\left(x+3\right)-x\left(x+3\right)=0\)
\(\Leftrightarrow x\left(x+3\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{0;-3;\dfrac{1}{2}\right\}\)
c) Ta có: \(x^2+\left(x+2\right)\left(11x-7\right)=4\)
\(\Leftrightarrow x^2+11x^2-7x+22x-14-4=0\)
\(\Leftrightarrow12x^2+15x-18=0\)
\(\Leftrightarrow12x^2+24x-9x-18=0\)
\(\Leftrightarrow12x\left(x+2\right)-9\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(12x-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\12x-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\12x=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{3}{4}\end{matrix}\right.\)
Vậy: \(S=\left\{-2;\dfrac{3}{4}\right\}\)
Bài `1:`
`h)(3/4x-1)(5/3x+2)=0`
`=>[(3/4x-1=0),(5/3x+2=0):}=>[(x=4/3),(x=-6/5):}`
______________
Bài `2:`
`b)3x-15=2x(x-5)`
`<=>3(x-5)-2x(x-5)=0`
`<=>(x-5)(3-2x)=0<=>[(x=5),(x=3/2):}`
`d)x(x+6)-7x-42=0`
`<=>x(x+6)-7(x+6)=0`
`<=>(x+6)(x-7)=0<=>[(x=-6),(x=7):}`
`f)x^3-2x^2-(x-2)=0`
`<=>x^2(x-2)-(x-2)=0`
`<=>(x-2)(x^2-1)=0<=>[(x=2),(x^2=1<=>x=+-2):}`
`h)(3x-1)(6x+1)=(x+7)(3x-1)`
`<=>18x^2+3x-6x-1=3x^2-x+21x-7`
`<=>15x^2-23x+6=0<=>15x^2-5x-18x+6=0`
`<=>(3x-1)(5x-1)=0<=>[(x=1/3),(x=1/5):}`
`j)(2x-5)^2-(x+2)^2=0`
`<=>(2x-5-x-2)(2x-5+x+2)=0`
`<=>(x-7)(3x-3)=0<=>[(x=7),(x=1):}`
`w)x^2-x-12=0`
`<=>x^2-4x+3x-12=0`
`<=>(x-4)(x+3)=0<=>[(x=4),(x=-3):}`
`m)(1-x)(5x+3)=(3x-7)(x-1)`
`<=>(1-x)(5x+3)+(1-x)(3x-7)=0`
`<=>(1-x)(5x+3+3x-7)=0`
`<=>(1-x)(8x-4)=0<=>[(x=1),(x=1/2):}`
`p)(2x-1)^2-4=0`
`<=>(2x-1-2)(2x-1+2)=0`
`<=>(2x-3)(2x+1)=0<=>[(x=3/2),(x=-1/2):}`
`r)(2x-1)^2=49`
`<=>(2x-1-7)(2x-1+7)=0`
`<=>(2x-8)(2x+6)=0<=>[(x=4),(x=-3):}`
`t)(5x-3)^2-(4x-7)^2=0`
`<=>(5x-3-4x+7)(5x-3+4x-7)=0`
`<=>(x+4)(9x-10)=0<=>[(x=-4),(x=10/9):}`
`u)x^2-10x+16=0`
`<=>x^2-8x-2x+16=0`
`<=>(x-2)(x-8)=0<=>[(x=2),(x=8):}`
1: \(\Leftrightarrow2x^2-10x-3x-2x^2=0\)
=>-13x=0
=>x=0
2: \(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
=>3x=13
=>x=13/3
3: \(\Leftrightarrow4x^4-6x^3-4x^3+6x^3-2x^2=0\)
=>-2x^2=0
=>x=0
4: \(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
=>-8x=6-14=-8
=>x=1
`1)2x(x-5)-(3x+2x^2)=0`
`<=>2x^2-10x-3x-2x^2=0`
`<=>-13x=0`
`<=>x=0`
___________________________________________________
`2)x(5-2x)+2x(x-1)=13`
`<=>5x-2x^2+2x^2-2x=13`
`<=>3x=13<=>x=13/3`
___________________________________________________
`3)2x^3(2x-3)-x^2(4x^2-6x+2)=0`
`<=>4x^4-6x^3-4x^4+6x^3-2x^2=0`
`<=>x=0`
___________________________________________________
`4)5x(x-1)-(x+2)(5x-7)=0`
`<=>5x^2-5x-5x^2+7x-10x+14=0`
`<=>-8x=-14`
`<=>x=7/4`
___________________________________________________
`5)6x^2-(2x-3)(3x+2)=1`
`<=>6x^2-6x^2-4x+9x+6=1`
`<=>5x=-5<=>x=-1`
___________________________________________________
`6)2x(1-x)+5=9-2x^2`
`<=>2x-2x^2+5=9-2x^2`
`<=>2x=4<=>x=2`
a) \(\left(x-4\right)^4+\left(x-2\right)^4=82\)
Đặt \(x-3=a\), ta có:
\(\left(a-1\right)^4+\left(a+1\right)^4=82\)
\(\Rightarrow\left[\left(a-1\right)^2\right]^2+\left[\left(a+1\right)^2\right]^2=82\)
\(\Rightarrow\left(a^2-2a+1\right)^2+\left(a+2a+1\right)^2=82\)
\(\Rightarrow\left(a^2+1\right)^2-4a\left(a^2+1\right)+4a^2+\left(a^2+1\right)^2+4a\left(a^2+1\right)+4a^2=82\)
\(\Rightarrow2\left[\left(a^2+1\right)^2+4a^2\right]=82\)
\(\Rightarrow\left(a^2+1\right)^2+4a^2=41\)
\(\Rightarrow a^4+2a^2+1+4a^2=41\)
\(\Rightarrow a^4+6a^2+1=41\)
\(\Rightarrow a^4+6a^2-40=0\)
\(\Rightarrow a^4-4a^2+10a^2-40=0\)
\(\Rightarrow a^2\left(a^2-4\right)+10\left(a^2-4\right)=0\)
\(\Rightarrow\left(a^2-4\right)\left(a^2+10\right)=0\)
\(\Rightarrow\left(a-2\right)\left(a+2\right)\left(a^2+10\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a-2=0\\a+2=0\\a^2+10=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}a=2\\a=-2\\a^2=-10\end{matrix}\right.\)
Vì a2 = -10 ( Không tồn tại )
\(\Rightarrow\left[{}\begin{matrix}a=2\\a=-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-3=2\\x-3=-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)
b)Sửa đề
\(6x^4+5x^3-38x^2+5x+6=0\)
Xét x = 0 không phải là nghiệm của phương trình
Xét \(x\ne0\), chia cả hai vế cho x2, ta được
\(6x^2+5x-38+\dfrac{5}{x}+\dfrac{6}{x^2}=0\)
\(\Rightarrow6\left(x^2+\dfrac{1}{x^2}\right)+5\left(x+\dfrac{1}{x}\right)-38=0\)
Đặt \(x+\dfrac{1}{x}=a\left(1\right)\)
\(\Rightarrow\left(x+\dfrac{1}{x}\right)^2=a^2\)
\(\Rightarrow x^2+\dfrac{1}{x^2}+2=a^2\)
\(\Rightarrow x^2+\dfrac{1}{x^2}=a^2-2\left(2\right)\)
Thay (1) và (2) vào phương trình
\(6\left(a^2-2\right)+5a-38=0\)
\(\Rightarrow6a^2-12+5a-38=0\)
\(\Rightarrow6a^2+5a-50=0\)
\(\Rightarrow6a^2-15a+20a-50=0\)
\(\Rightarrow3a\left(2a-5\right)+10\left(2a-5\right)=0\)
\(\Rightarrow\left(2a-5\right)\left(3a+10\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2a-5=0\\3a+10=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2a=5\\3a=-10\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}a=\dfrac{5}{2}\\a=-\dfrac{10}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{x}=\dfrac{5}{2}\\x+\dfrac{1}{x}=-\dfrac{10}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{2}\\x=-\dfrac{1}{3}\\x=-3\end{matrix}\right.\)
\(b,\)\(\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}\)
\(\Rightarrow\left(\frac{x+1}{2008}+1\right)+\left(\frac{x+2}{2007}+1\right)+\left(\frac{x+3}{2006}+1\right)=\left(\frac{x+4}{2005}+1\right)+\left(\frac{x+5}{2004}+1\right)+\left(\frac{x+6}{2003}+1\right)\)
\(\Rightarrow\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}=\frac{x+2009}{2005}+\frac{x+2009}{2004}+\frac{x+2009}{2003}\)
\(\Rightarrow\left(x+9\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}\right)=\left(x+9\right)\left(\frac{1}{2005}+\frac{1}{2004}+\frac{1}{2003}\right)\)
\(\Rightarrow\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}=\frac{1}{2005}+\frac{1}{2004}+\frac{1}{2003}\left(KTM\right)\)
\(\text{Giải}\)
\(b,\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}\)
\(\Leftrightarrow\left(x+2009\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)=0\)
\(\Leftrightarrow x+2009=0\Leftrightarrow x=-2009\)
a: Ta có: \(3x-\left(3x+2\right)=x+3\)
\(\Leftrightarrow x+3=-2\)
hay x=-5
b: Ta có: \(\dfrac{5x-1}{4}+\dfrac{2x-1}{3}=\dfrac{3x}{2}\)
\(\Leftrightarrow15x-3+8x-4=18x\)
\(\Leftrightarrow5x=7\)
hay \(x=\dfrac{7}{5}\)
Đang rảnh, buồn ngủ nên giải cho tỉnh táo :D
Ta nhận thấy x=0 không phải là nghiệm của phương trình, vậy ta chia cả 2 vế của phương trình cho x2 khác 0, ta được:
\(6x^2+5x-38+\dfrac{5}{x}+\dfrac{6}{x^2}=0\)
\(\Leftrightarrow6\left(x^2+\dfrac{1}{x}\right)+5\left(x+\dfrac{1}{x}\right)-38=0\)
Đặt \(x+\dfrac{1}{x}=y\Rightarrow x^2+\dfrac{1}{x^2}=y^2-2\)
Ta được: \(6\left(y^2-2\right)+5y-38=0\)
Do đó: y1=2,5;y2=-10/3
Với y=2,5\(\Rightarrow x+\dfrac{1}{x}=2,5\Rightarrow x_1=2;x_2=0,5\)
Với y=-10/3
\(\Rightarrow x+\dfrac{1}{x}=-\dfrac{10}{3}\Rightarrow x_3=-\dfrac{1}{3};x_4=-3\)
Vậy: \(S=\left\{2;0,5;-\dfrac{1}{3};-3\right\}\)
Bài a tự giải
Bài b thì biến đổi xong rồi đặt ẩn phụ \(y=x+\dfrac{1}{x}\)
Bài c:
Đặt x-1=y
Phương trình trở thành: \(\left(y+2\right)^4+\left(y-2\right)^4=82\)
Rút gọn ta được: \(2y^4+48y^2-50=0\)
Đặt y2=z ( \(z\ge0\) )
Phương trình này cho z1=1, z2=-25(Loại)
\(z=1\Rightarrow y^2=1\Rightarrow y=\pm1\)
\(\Rightarrow x_1=2;x_2=0\)