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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
Rút gọn hết ta được :
a/ 41x - 17 = -21
=> 41x = -4 => x = 4/41
b/ 34x - 17 = 0
=> 34x = 17
=> x = 17/34 = 1/2
c/ 19x + 56 = 52
=> 19x = -4
=> x = -4/19
d/ 20x2 - 16x - 34 = 10x2 + 3x - 34
=> 10x2 - 19x = 0
=> x(10x - 19) = 0
=> x = 0
hoặc 10x - 19 = 0 => 10x = 19 => x = 19/10
Vậy x = 0 ; x = 19/10
Rút gọn hết ta được :
a/ 41x - 17 = -21
=> 41x = -4 => x = 4/41
b/ 34x - 17 = 0
=> 34x = 17
=> x = 17/34 = 1/2
c/ 19x + 56 = 52
=> 19x = -4
=> x = -4/19
d/ 20x 2 - 16x - 34 = 10x 2 + 3x - 34
=> 10x 2 - 19x = 0
=> x(10x - 19) = 0
=> x = 0 hoặc 10x - 19 = 0
=> 10x = 19
=> x = 19/10
Vậy x = 0 ; x = 19/10
a) (2x+3)(x-4)+(x-5)(x-2)=(3x-5)(x-4)
(2x+3)(x-4)+(x-5)(x-2)-(3x-5)(x-4)=0\(2x^2-8x+3x-12+x^2-2x-5x+10-3x^2+12x+5x-20\)=0
(\(2x^2+x^2-3x^2\))+(-8x+3x-2x-5x+12x+5x)+(-12+10-20) =0 5x-22 =0
5x = 22
x = \(\dfrac{22}{5}\)
Vậy x= \(\dfrac{22}{5}\)
b) (8x-3)(3x+2)-(4x+7)(x+4)=(2x+1)(5x-1)
(8x-3)(3x+2)-(4x+7)(x+4)-(2x+1)(5x-1)=0
\(24x^2\) +16x-9x-6\(-4x^2\) -16x-7x-28\(-10x^2\) +2x-5x+1=0
(24\(x^2-4x^2-10x^2\))+(16x-9x-16x-7x+2x-5x)+(-6-28+1)=0
10\(x^2-19x-33\)=0
10\(x^2+11x-30x-33=0\)
x(10x+11)-3(10x+11)=0
(x-3) (10x+11)=0
=>x-3=0 => x=3 =>x=3
10x+11=0 10x=-11 x=\(\dfrac{-11}{10}\)
Vậy x=3 hoặc x=\(\dfrac{-11}{10}\)
Tìm x, biết:
a) (2x+3)(x-4)+(x-5)(x-2)=(3x-5)(x-4)
<=> \(2x^2-8x+3x-12+x^2-2x-5x+10=3x^2-12x-5x+20\)
<=> \(2x^2-8x+3x+x^2-2x-5x-3x^2+12x+5x=12-10+20\)
<=> \(5x=22\)
<=> \(x=\dfrac{22}{5}\)
Vậy \(S=\left\{\dfrac{22}{5}\right\}\)
Tìm min:
$F=3x^2+x-2=3(x^2+\frac{x}{3})-2$
$=3[x^2+\frac{x}{3}+(\frac{1}{6})^2]-\frac{25}{12}$
$=3(x+\frac{1}{6})^2-\frac{25}{12}\geq \frac{-25}{12}$
Vậy $F_{\min}=\frac{-25}{12}$. Giá trị này đạt tại $x+\frac{1}{6}=0$
$\Leftrightarrow x=\frac{-1}{6}$
Tìm min
$G=4x^2+2x-1=(2x)^2+2.2x.\frac{1}{2}+(\frac{1}{2})^2-\frac{5}{4}$
$=(2x+\frac{1}{2})^2-\frac{5}{4}\geq 0-\frac{5}{4}=\frac{-5}{4}$ (do $(2x+\frac{1}{2})^2\geq 0$ với mọi $x$)
Vậy $G_{\min}=\frac{-5}{4}$. Giá trị này đạt tại $2x+\frac{1}{2}=0$
$\Leftrightarrow x=\frac{-1}{4}$
một đòn bẫy dài một mét .đặt ở đâu để có thể dùng 3600n có thể nâng tảng đá nặng 120kg?
\(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)\)
\(\Leftrightarrow24x^2+7x-6-\left(4x^2+23x+28\right)=10x^2+3x-1\)
\(\Leftrightarrow20x^2-16x-34=10x^2+3x-1\)
\(\Leftrightarrow10x^2-19x-33=0\)
\(\Leftrightarrow10x^2-30x+11x-33=0\)
\(\Leftrightarrow10x\left(x-3\right)+11\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(10x+11\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\10x+11=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-\frac{11}{10}\end{cases}}\)
Vậy \(x\in\left\{3;-\frac{11}{10}\right\}.\)
Bài làm :
Ta có :
\(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)\)
\(\Leftrightarrow24x^2+7x-6-\left(4x^2+23x+28\right)=10x^2+3x-1\)
\(\Leftrightarrow20x^2-16x-34=10x^2+3x-1\)
\(\Leftrightarrow10x^2-19x-33=0\)
\(\Leftrightarrow10x^2-30x+11x-33=0\)
\(\Leftrightarrow10x\left(x-3\right)+11\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(10x+11\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\10x+11=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-\frac{11}{10}\end{cases}}\)
Vậy x=3 hoặc x=-11/10