(x-7) (x+7)+8x^2=(3x-2)^2-4x
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ĐK : x khác 7 ; 1
\(\frac{4x}{x^2-8x+7}+\frac{3x}{4x^2-10x+7}=1\)
\(\Leftrightarrow\frac{4x}{\left(x-7\right)\left(x-1\right)}+\frac{3x}{4x^2-10x+7}=1\)
\(\Leftrightarrow\frac{4x\left(4x^2-10x+7\right)}{\left(x-7\right)\left(x-1\right)\left(4x^2-10x+7\right)}+\frac{3x\left(x-7\right)\left(x-1\right)}{\left(x-1\right)\left(x-7\right)\left(4x^2-10x+7\right)}=1\)
\(\Leftrightarrow16x^3-40x^2+28x+3x^3-24x^2+21x=1\)
\(\Leftrightarrow19x^3-64x^2+49x-1=0\) vô nghiệm
Đề ko sai ak :)? từ cái chỗ 4x^2 - 10x + 7 ý
\(P\left(x\right)=8x^3\) + 5x -1
+ \(Q\left(x\right)\)= \(4x^2\) - 3x + 7
+ \(R\left(x\right)=8x^3+8x^2+7x\)
Tổng : 16x^3 + 12x^2 +9x + 6
một đòn bẫy dài một mét .đặt ở đâu để có thể dùng 3600n có thể nâng tảng đá nặng 120kg?
`@` `\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`
Đk: \(x\ge-2\)
PT \(\Leftrightarrow\) \(x\left(12-4\sqrt{x+2}\right)+3x^2-20x-7=0\)
\(\Leftrightarrow x.\dfrac{144-16\left(x+2\right)}{12+4\sqrt{x+2}}+\left(x-7\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\dfrac{-4x\left(x-7\right)}{3+\sqrt{x+2}}+\left(x-7\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\\\left(3+\sqrt{x+2}\right)\left(3x+1\right)=4x\end{matrix}\right.\)
Đặt \(u=\sqrt{x+2}\Leftrightarrow x=u^2-2\left(u\ge0\right)\)
PT (2) \(\Leftrightarrow\left(3+u\right)\left(3u^2-5\right)=4\left(u^2-2\right)\)
\(\Leftrightarrow9u^2-15+3u^3-5u=4u^2-8\)
\(\Leftrightarrow3u^3+5u^2-5u-7=0\) \(\Leftrightarrow u=\dfrac{-1+\sqrt{22}}{3}\)
\(\Leftrightarrow x=\dfrac{5-2\sqrt{22}}{9}\)
Vậy...
Lời giải:
ĐKXĐ: $x\geq -2$
PT $\Leftrightarrow 3x^2-20x-7=4x\sqrt{x+2}-12x$
$\Leftrightarrow (x-7)(3x+1)=4x(\sqrt{x+2}-3)=4x.\frac{x-7}{\sqrt{x+2}+3}$
$\Leftrightarrow x-7=0$ hoặc $3x+1=\frac{4x}{\sqrt{x+2}+3}$
Nếu $x-7=0\Leftrightarrow x=7$ (tm)
Nếu $3x+1=\frac{4x}{\sqrt{x+2}+3}$
$\Leftrightarrow 9x+3+(3x+1)\sqrt{x+2}=4x$
$\Leftrightarrow 5x+3+(3x+1)\sqrt{x+2}=0$
$\Leftrightaqrrow 5x+3=-(3x+1)\sqrt{x+2}$
$\Rightarrow (5x+3)^2=(3x+1)^2(x+2)$
$\Leftrightarrow 9x^3-x^2-17x-7=0$
$\Leftrightarrow (x+1)(9x^2-10x-7)=0$
$\Rightarrow$........
8x2+30x+7=0
8x2+16x+14x+7=0
8x(x+2) +7(x+2)=0
(8x+7)(x+2)=0
=>\(\orbr{\begin{cases}8x+7=0\\x+2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-\frac{7}{8}\\x=-2\end{cases}}}\)
a) \(\left(x+2\right)\left(x+3\right)-\left(x+1\right)\left(x+7\right)=6\)
\(\Leftrightarrow x^2+5x+6-x^2-8x-7=6\)
\(\Leftrightarrow-3x=7\)
\(\Leftrightarrow x=-\frac{7}{3}\)
b) \(\left(8x-3\right)\left(3x+2\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)-33\)
\(\Leftrightarrow\left(8x-3\right)\left(9x^2+12x+4\right)-4x^2-23x-28=10x^2+3x-1-33\)
\(\Leftrightarrow72x^3+69x^2-4x-12-14x^2-26x+6=0\)
\(\Leftrightarrow72x^3+55x^2-30x-6=0\)
Nghiệm vô tỉ: \(x_1=-1,078...\) ; \(x_2=0,476...\) ; \(x_3=-0,162...\)
a) (x + 2)(x + 3) - (x + 1)(x + 7) = 6
=> x(x + 3) + 2(x + 3) - x(x + 7) - 1(x + 7) = 6
=> x2 + 3x + 2x + 6 - x2 - 7x - x - 7 = 6
=> x2 + 5x + 6 - x2 - 7x - x - 7 = 6
=> (x2 - x2) + (5x - 7x - x) + (6 - 7) = 6
=> -3x - 1 = 6
=> -3x = 7
=> x = -7/3
b) (8x - 3)(3x + 2)(3x + 2) - (4x + 7)(x + 4) = (2x + 1)(5x - 1) - 33
=> (8x - 3)(9x2 + 12x + 4) - [4x(x + 4) + 7(x + 4)] = 2x(5x - 1) + 1(5x - 1) - 33
=> 8x(9x2 + 12x + 4) - 3(9x2 + 12x + 4) - (4x2 + 16x + 7x + 28) = 10x2 - 2x + 5x - 1 - 33
=> 72x3 + 96x2 + 32x - 27x2 - 36x - 12 - 4x2 - 16x - 7x - 28 - 10x2 + 2x - 5x + 1 + 33 = 0
=> 72x3 + (96x2 - 27x2 - 10x2 - 4x2) + (32x - 36x - 16x - 7x + 2x - 5x) + (-12 - 28 + 1 + 33) = 0
=> 72x3 + 55x2 - 30x - 6 = 0
=> x vô nghiệm
\(\left(x-7\right)\left(x+7\right)+8x^2=\left(3x-2\right)^2-4x\)
\(\Leftrightarrow\left(x^2-49\right)+8x^2=\left(9x^2-12x+4\right)-4x\)
\(\Leftrightarrow9x^2-49=9x^2-16x+4\)
\(\Leftrightarrow-16x=-53\)
\(\Leftrightarrow x=\dfrac{53}{16}\)