(4x - 1)^4 = (4x -1 )^3
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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 \(\sqrt{x+9}=t\left(t\ge0\right)\Rightarrow9=t^2-x\)
\(\left(1\right)\Leftrightarrow t+x^2+x-t^2=0\)
\(\Leftrightarrow\left(x+t\right)\left(x-t+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-t\\x=t-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{x+9}\\x=\sqrt{x+9}-1\end{matrix}\right.\)
\(\Leftrightarrow...\)
1. \(\sqrt{x^2-4x+3}=x-2\)
<=> x2 - 4x + 3 = (x - 2)2
<=> x2 - 4x + 3 = x2 - 4x + 4
<=> x2 - x2 - 4x + 4x = 1
<=> 0 = 1 (Vô lí)
vậy PT có nghiệm là S = \(\varnothing\)
2. \(\sqrt{4x^2-4x+1}=x-1\)
<=> \(\sqrt{\left(2x-1\right)^2}=x-1\)
<=> 2x - 1 = x - 1
<=> 2x - x = -1 + 1
<=> x = 0
a: Ta có: \(\left(3x-1\right)^2-\left(3x+4\right)\left(3x-4\right)=32\)
\(\Leftrightarrow9x^2-6x+1-9x^2+16=32\)
\(\Leftrightarrow-6x=15\)
hay \(x=-\dfrac{5}{2}\)
b: Ta có: \(\left(4x+3\right)^2-\left(4x-1\right)\left(4x+1\right)=-14\)
\(\Leftrightarrow16x^2+24x+9-16x^2+1=-14\)
\(\Leftrightarrow24x=-24\)
hay x=-1
\(a/\)
\(4x-4y+x^2-2xy+y^2\)
\(=\left(4x-4y\right)+\left(x^2-2xy+y^2\right)\)
\(=4\left(x-y\right)+\left(x-y\right)^2\)
\(=\left(x-y\right)\left(4+x-y\right)\)
\(b/\)
\(x^4-4x^3-8x^2+8x\)
\(=\left(x^4+8x\right)-\left(4x^3+8x^2\right)\)
\(=x\left(x^3+8\right)-4x^2\left(x+2\right)\)
\(=x\left(x+2\right)\left(x^2-2x+4\right)-4x^2\left(x+2\right)\)
\(=x\left(x+2\right)\left(x^2-2x+4-4x\right)\)
\(=x\left(x+2\right)\left(x^2-6x-4\right)\)
\(d/\)
\(x^4-x^2+2x-1\)
\(=x^4-\left(x-1\right)^2\)
\(=\left(x^2+x-1\right)\left(x^2-x+1\right)\)
\(e/\)(Xem lại đề)
\(x^4+x^3+x^2+2x+1\)
\(=\left(x^4+x^3\right)+\left(x^2+2x+1\right)\)
\(=x^3\left(x+1\right)+\left(x+1\right)^2\)
\(=\left(x+1\right)\left(x^3+x+1\right)\)
\(f/\)
\(x^3-4x^2+4x-1\)
\(=x\left(x^2-4x+4\right)-1^2\)
\(=x\left(x-2\right)^2-1\)
\(=[\sqrt{x}\left(x-2\right)]^2-1\)
\(=[\sqrt{x}\left(x-2\right)-1][\sqrt{x}\left(x-2\right)+1]\)
\(c/\)
\(x^3+x^2-4x-4\)
\(=\left(x^3-2x^2\right)+\left(3x^2-6x\right)+\left(2x-4\right)\)
\(=x^2\left(x-2\right)+3x\left(x-2\right)+2\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+3x+2\right)\)
\(=\left(x-2\right)[\left(x^2+x\right)+\left(2x+2\right)]\)
\(=\left(x-2\right)\left(x+1\right)\left(x+2\right)\)
1: =>x^2-x=3-x
=>x^2=3
=>x=căn 3 hoặc x=-căn 3
2: =>x^2-4x+3=x^2-4x+4 và x>=2
=>3=4(vô lý)
3: =>2|x-1|=6
=>|x-1|=3
=>x-1=3 hoặc x-1=-3
=>x=-2 hoặc x=4
4: =>|2x-3|=|x-2|
=>2x-3=x-2 hoặc 2x-3=-x+2
=>x=1 hoặc x=5/3
5: =>\(\sqrt{x+2}\left(\sqrt{x-2}+\sqrt{x+2}\right)=0\)
=>x+2=0
=>x=-2
`#3107`
`a)`
`A=`\(3x^4 + \dfrac{1}3xyz - 3x^4 - \dfrac{4}3xyz + 2x^2y - 6z\)
`= (3x^4 - 3x^4) + (1/3xyz - 4/3xyz) + 2x^2y - 6z`
`= -xyz + 2x^2y - 6z`
Thay `x = 1; y = 3` và `z = 1/3` vào A
`A = -1*3*1/3 + 2*1^2*3 - 6*1/3`
`= -1 + 6 - 2`
`= 6 - 3`
`= 3`
Vậy, `A=3`
`b)`
`B=`\(4x^3 - \dfrac{2}7xyz - 4x^3 - \dfrac{4}3xyz + 4x^2y\)
`= (4x^3 - 4x^3) + (-2/7xyz - 4/3xyz) + 4x^2y`
`= -34/21 xyz + 4x^2y`
Thay `x = -1; y = 2` và `z = -1/2` vào B
`B = -34/21*(-1)*2*(-1/2) + 4*(-1)^2 * 2`
`= -34/21 + 8`
`= 134/21`
Vậy, `B = 134/21`
`c)`
`C=`\(4x^2 + \dfrac{1}2xyz - \dfrac{2}3xy^2z - 5x^2yz + \dfrac{3}4xyz\)
`= 4x^2 + (1/2xyz + 3/4xyz) - 2/3xy^2z - 5x^2yz `
`= 4x^2 + 5/4xyz - 2/3xy^2z - 5x^2yz`
Ta có:
`|y| = 2`
`=> y = +-2`
Thay `x = -1; y = 2` và `z = 1/2` vào C
`4*(-1)^2 + 5/4*(-1)*2*1/2 - 2/3*(-1)*2^2*1/2 - 5*(-1)^2*2*1/2`
`= 4 - 5/4 + 4/3 - 5`
`= -11/12`
Vậy, với `x = -1; y = 2; z = 1/2` thì `B = -11/12`
Thay `x = -1; y = -2; z = 1/2`
`B = 4*(-1)^2 + 5/4*(-1)*(-2)*1/2 - 2/3*(-1)*(-2)^2*1/2 - 5*(-1)^2*(-2)*1/2`
`= 4 + 5/4 + 4/3 + 5`
`= 139/12`
Vậy, với `x = -1; y = -2; z = 1/2` thì `B = 139/12.`
\(\left|2+3x\right|=\left|4x-3\right|\)
\(\Leftrightarrow\orbr{\begin{cases}2+3x=4x-3\\2+3x=3-4x\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=5\\x=\frac{1}{7}\end{cases}}\)
Vậy \(x\in\left\{\frac{1}{7};5\right\}\)
\(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\\frac{3}{2}x+\frac{1}{2}=1-4x\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{5}\\x=\frac{1}{11}\end{cases}}\)
Vậy \(x\in\left\{\frac{1}{11};\frac{3}{5}\right\}\)
\(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
\(\Leftrightarrow\left|\frac{5}{4}x-\frac{7}{2}\right|=\left|\frac{5}{8}x+\frac{3}{5}\right|\)
Giải tiếp tương tự
Sau đó giải tiếp câu còn lại
đặt 4x-1=t
=>t4=t3
=>t4-t3=0
=>t3(t-1)=0
=>t3=0 hoặc t-1=0
=>t=0;1
=>x=1/4;1/2