(8x-7)(8x-5)(2x-1)(4x-1)=9
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\(\left(8x-7\right)\left(8x-5\right)\left(2x-1\right)\left(4x-1\right)=9\)
\(\Leftrightarrow\left(8x-7\right)\left(8x-5\right)\left(4x-4\right)\left(8x-2\right)=72\)
Đặt a = 8x - 5, ta được:
\(\left(a-2\right).a\left(a+1\right)\left(a+3\right)=72\)
\(\Leftrightarrow a^4+4a^3+3a^2-2a^3-8a^2-6a-72=0\)
\(\Leftrightarrow a^4+4a^3-2a^3-8a^2+3a^2+12a-18a-72=0\)\(\Leftrightarrow\left(a^4+4a^3\right)-\left(2a^3+8a^2\right)+\left(3a^2+12a\right)-\left(18a+72\right)=0\)
\(\Leftrightarrow a^3\left(a+4\right)-2a^2\left(a+4\right)+3a\left(a+4\right)-18\left(a+4\right)=0\) \(\Leftrightarrow\left(a+4\right)\left(a^3-2a^2+3a-18\right)=0\)
\(\Leftrightarrow\left(a+4\right)\left(a^3-3a^2+a^2-3a+6a-18\right)=0\)
\(\Leftrightarrow\left(a+4\right)\left[\left(a^3-3a^2\right)+\left(a^2-3a\right)+\left(6a-18\right)\right]=0\)
\(\Leftrightarrow\left(a+4\right)\left[a^2\left(a-3\right)+a\left(a-3\right)+6\left(a-3\right)\right]=0\)
\(\Leftrightarrow\left(a+4\right)\left(a-3\right)\left(a^2+a+6\right)=0\)
Ta có: \(a^2+a+6=a^2+2.a.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{23}{4}=\left(a+\dfrac{1}{2}\right)^2+\dfrac{23}{4}\)
\(\left(a+\dfrac{1}{2}\right)\ge0\)
Suy ra \(\left(a+\dfrac{1}{2}\right)^2+\dfrac{23}{4}>0\)
=> a2 +a+6 = 0 (loại)
Suy ra: a = -4 hoặc a=3
Với a = -4, ta được:
8x - 5 = -4
=> x = \(\dfrac{1}{8}\)
Với a = 3, ta được:
8x - 5 = 3
=> x = 1
a) \(\left(8x+5\right)^2\left(4x+3\right)\left(2x+1\right)=9\)
\(\Leftrightarrow\left(64x^2+8x+25\right)\left(8x^2+10x+3\right)-9=0\)
Đặt a = \(8x^2+10x+3\)
\(\left(8a+1\right)a-9=0\)
\(\Leftrightarrow8a^2+a-9=0\)
\(\Leftrightarrow\left(a-1\right)\left(8a+9\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=1\\a=-\frac{9}{8}\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}8x^2+10x+3=1\\8x^2+10x+3=-\frac{9}{8}\end{cases}}\)
mà \(8x^2+10x+3=1\Rightarrow8x^2+10x+2=0\)
\(\Rightarrow2\left(x+1\right)\left(4x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=-1\\x=-0,25\end{cases}}\)
Lời giải:
PT \(\Leftrightarrow [(8x-7)(4x-1)][(8x-5)(2x-1)]=9\)
\(\Leftrightarrow (32x^2-36x+7)(16x^2-18x+5)=9\)
Đặt \(16x^2-18x=a\). PT trở thành:
\((2a+7)(a+5)=9\)
\(\Leftrightarrow 2a^2+17a+26=0\)
\(\Rightarrow \left[\begin{matrix} a=-2\\ a=\frac{-13}{2}\end{matrix}\right.\)
\(a=-2\Leftrightarrow 16x^2-18x+2=0\)
\(\Leftrightarrow (8x-1)(x-1)=0\Rightarrow \left[\begin{matrix} x=1\\ x=\frac{1}{8}\end{matrix}\right.\)
\(a=\frac{-13}{2}\Leftrightarrow 16x^2-18x+\frac{13}{2}=0\) (pt vô nghiệm)
Vậy............
\(\dfrac{x-1}{2x^2-4x}-\dfrac{7}{8x}=\dfrac{5-x}{4x^2-8x}-\dfrac{1}{8x-16}\) ( ĐKXĐ: \(x\ne0;x\ne2\) )
\(\Leftrightarrow\dfrac{x-1}{2x\left(x-2\right)}-\dfrac{7}{8x}=\dfrac{5-x}{4x\left(x-2\right)}-\dfrac{1}{8\left(x-2\right)}\)
\(\Leftrightarrow\dfrac{\left(x-1\right)4}{8x\left(x-2\right)}-\dfrac{7\left(x-2\right)}{8x\left(x-2\right)}=\dfrac{2\left(5-x\right)}{8x\left(x-2\right)}-\dfrac{1x}{8x\left(x-2\right)}\)
\(\Rightarrow4x-4-7x+14=10-2x-x\)
\(\Leftrightarrow-3x+2x+x=10+4-14\)
\(\Leftrightarrow0=0\)
Vậy pt đã cho có nghiệm đúng với mọi x
Trả lời:
\(\frac{x-1}{2x^2-4x}-\frac{7}{8x}=\frac{5-x}{4x^2-8x}-\frac{1}{8x-16}\)\(\left(đkxđ:x\ne0;x\ne2\right)\)
\(\Leftrightarrow\frac{x-1}{2x\left(x-2\right)}-\frac{7}{8x}=\frac{5-x}{4x\left(x-2\right)}-\frac{1}{8\left(x-2\right)}\)
\(\Leftrightarrow\frac{4\left(x-1\right)}{8x\left(x-2\right)}-\frac{7\left(x-2\right)}{8x\left(x-2\right)}=\frac{2\left(5-x\right)}{8x\left(x-2\right)}-\frac{x}{8x\left(x-2\right)}\)
\(\Rightarrow4\left(x-1\right)-7\left(x-2\right)=2\left(5-x\right)-x\)
\(\Leftrightarrow4x-4-7x+14=10-2x-x\)
\(\Leftrightarrow10-3x=10-3x\)
\(\Leftrightarrow-3x+3x=10-10\)
\(\Leftrightarrow0x=0\)( luôn thỏa mãn )
Vậy S = R với \(x\ne0;x\ne2\)
`@` `\text {Ans}`
`\downarrow`
`a)`
`3x(4x-1) - 2x(6x-3) = 30`
`=> 12x^2 - 3x - 12x^2 + 6x = 30`
`=> 3x = 30`
`=> x = 30 \div 3`
`=> x=10`
Vậy, `x=10`
`b)`
`2x(3-2x) + 2x(2x-1) = 15`
`=> 6x- 4x^2 + 4x^2 - 2x = 15`
`=> 4x = 15`
`=> x = 15/4`
Vậy, `x=15/4`
`c)`
`(5x-2)(4x-1) + (10x+3)(2x-1) = 1`
`=> 5x(4x-1) - 2(4x-1) + 10x(2x-1) + 3(2x-1)=1`
`=> 20x^2-5x - 8x + 2 + 20x^2 - 10x +6x - 3 =1`
`=> 40x^2 -17x - 1 = 1`
`d)`
`(x+2)(x+2)-(x-3)(x+1)=9`
`=> x^2 + 2x + 2x + 4 - x^2 - x + 3x + 3=9`
`=> 6x + 7 =9`
`=> 6x = 2`
`=> x=2/6 =1/3`
Vậy, `x=1/3`
`e)`
`(4x+1)(6x-3) = 7 + (3x-2)(8x+9)`
`=> 24x^2 - 12x + 6x - 3 = 7 + (3x-2)(8x+9)`
`=> 24x^2 - 12x + 6x - 3 = 7 + 24x^2 +11x - 18`
`=> 24x^2 - 6x - 3 = 24x^2 + 18x -11`
`=> 24x^2 - 6x - 3 - 24x^2 + 18x + 11 = 0`
`=> 12x +8 = 0`
`=> 12x = -8`
`=> x= -8/12 = -2/3`
Vậy, `x=-2/3`
`g)`
`(10x+2)(4x- 1)- (8x -3)(5x+2) =14`
`=> 40x^2 - 10x + 8x - 2 - 40x^2 - 16x + 15x + 6 = 14`
`=> -3x + 4 =14`
`=> -3x = 10`
`=> x= - 10/3`
Vậy, `x=-10/3`
X=1
k nhé
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