tim x biet: (5x-2)(3x+1)+(7-15x)(x+3)=-20 Merci
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
(5x - 2) . (3x + 1) + (7 - 15x) . (x + 3) = 20
=> 15x2 + 5x - 6x - 2 + 7x + 21 - 15x2 - 45x = 20
=> (15x2 - 15x2) + (5x - 6x + 7x - 45x) = 20 + 2 - 21
=> 0 + [x . (5 - 6 + 7 - 45)] = 1
=> -39x = 1
=> x = 1 : (-39)
Vậy x = \(\frac{-1}{39}\)
\(b,4x^2-x-5=0\)
\(\Leftrightarrow4x^2-5x+4x-5=0\)
\(\Leftrightarrow x\left(4x-5\right)+4x-5=0\)
\(\Leftrightarrow\left(4x-5\right)\left(x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=-1\\x=\frac{5}{4}\end{cases}}\)
Bài 2
\(a,x^3+5x^2+3x-9\)
\(\Leftrightarrow x^3-x^2+6x^2-6x+9x-9\)
\(\Leftrightarrow x^2\left(x-1\right)+6x\left(x-1\right)+9\left(x-1\right)\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+6x+9\right)\)
\(\Leftrightarrow\left(x-1\right)\left(x+3\right)^2\)
b,\(x^3-7x-6\)
\(\Leftrightarrow x^3-3x^2+3x^2-9x+2x-6\)
\(\Leftrightarrow x^2\left(x-3\right)+3x\left(x-3\right)+2\left(x-3\right)\)
\(\Leftrightarrow\left(x-3\right)\left(x^2+3x+2\right)\)
\(\Leftrightarrow\left(x-3\right)\left(x+1\right)\left(x+2\right)\)
c,\(3x^3-7x^2+17x-5\)
\(\Leftrightarrow3x^3-x^2-6x^2+2x+15x-5\)
\(\Leftrightarrow x^2\left(3x-1\right)-2x\left(3x-1\right)+5\left(3x-1\right)\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2-2x+5\right)\)
Bài 1:
a)\(28x^3+15x^2+75x+125=0\)
\(\Leftrightarrow\left(4x+5\right)\left(7x^2-5x+25\right)=0\)
Dễ thấy: \(7x^2-5x+25=7\left(x-\frac{5}{14}\right)^2+\frac{675}{28}>0\)
\(\Rightarrow4x+5=0\Rightarrow x=-\frac{5}{4}\)
b)\(4x^2-x-5=0\)
\(\Leftrightarrow\left(x+1\right)\left(4x-5\right)=0\)
\(\Rightarrow x=-1;x=\frac{5}{4}\)
Bài 2:
a)\(x^3+5x^2+3x-9\)
\(=\left(x-1\right)\left(x+3\right)^2\)
b)\(x^3-7x-6\)
\(=\left(x-3\right)\left(x+1\right)\left(x+2\right)\)
c)\(3x^3-7x^2+17x-5\)
\(=\left(3x-1\right)\left(x^2-2x+5\right)\)
a) Ta có: \(\frac{3x+2}{5x+7}=\frac{3x-1}{5x+1}\)
\(\Leftrightarrow\left(3x+2\right)\left(5x+1\right)=\left(5x+7\right)\left(3x-1\right)\)
\(\Leftrightarrow3x\left(5x+1\right)+2\left(5x+1\right)=5x\left(3x-1\right)+7\left(3x-1\right)\)
\(\Leftrightarrow15x^2+3x+10x+2=15x^2-5x+21x-7\)
\(\Leftrightarrow15x^2-15x^2+3x+10x+5x-21x=-7-2\)
\(\Leftrightarrow-3x=-9\)
\(\Leftrightarrow x=3\)
Vậy x = 3
b) Ta có: \(\frac{x+1}{2x+1}=\frac{0,5x+2}{x+3}\Leftrightarrow\left(x+1\right)\left(x+3\right)=\left(2x+1\right)\left(0,5x+2\right)\)
\(\Leftrightarrow x\left(x+3\right)+\left(x+3\right)=2x\left(0,5x+2\right)+\left(0,5x+2\right)\)
\(\Leftrightarrow x^2+3x+x+3=x^2+4x+0,5x+2\)
\(\Leftrightarrow x^2-x^2+3x+x-4x-0,5x=2-3\)
\(\Leftrightarrow-0,5x=-1\Leftrightarrow x=2\)
Vậy x = 2
a/ (x-1)2-(4x+3)(2-x)=x2-2x+1-(8x-4x2+6-3x)
=x2-2x+1-8x+4x2-6+3x=5x2-7x-6
b/ (15x3y2 - 6x2y3) : 3x2y2 = 5x - 2y
c/ \(\dfrac{x+7}{x-7}-\dfrac{x-7}{x+7}+\dfrac{4x^2}{x^2-49}\)=\(\dfrac{\left(x+7\right)^2-\left(x-7\right)^2+4x^2}{\left(x-7\right)\left(x+7\right)}\)=\(\dfrac{x^2+14x+49-\left(x^2-14x+49\right)+4x^2}{\left(x-7\right)\left(x+7\right)}\)=\(\dfrac{28x+4x^2}{\left(x-7\right)\left(x+7\right)}\)=\(\dfrac{4x\left(x+7\right)}{\left(x-7\right)\left(x+7\right)}\)=\(\dfrac{4x}{x-7}\)
1/ \(5x-3\left\{4x-2\left[4x-\left(5x-2\right)\right]\right\}=182\) (1)
\(\Leftrightarrow5x-3\cdot\left[4x-2\left(4x-5x+2\right)\right]\)
\(\Leftrightarrow5x-3\cdot\left[4x-2\left(-x+2\right)\right]=182\)
\(\Leftrightarrow5x-3\left(4x+2x-4\right)=182\)
\(\Leftrightarrow5x-3\left(6x-4\right)=182\)
\(\Leftrightarrow5x-18x+12=182\)
\(\Leftrightarrow-13x+12=182\)
\(\Leftrightarrow-13x=182-12\)
\(\Leftrightarrow-13x=170\)
\(\Leftrightarrow x=-\dfrac{170}{13}\)
Vậy tập nghiệm phương trình (1) \(S=\left\{-\dfrac{170}{13}\right\}\)
2/ \(A=3x\left(5x^2-4\right)+x^2\left(8-15x\right)-8x^2\)
\(=15x^3-12+8x^2-15x^3-8x^2\)
\(=-12x\)
ta thấy \(\left|x\right|=3\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
Thay lần lượt các giá trị của x vào biểu thức A cho ta 2 kết quả là -36 tại x = 3 và 36 tại x = -3.
1,
\(5x-3\left\{4x-2\left[4x-\left(5x-2\right)\right]\right\}=182\Leftrightarrow5x-3\left[4x-2\left(2-x\right)\right]=182\)\(\Leftrightarrow5x-3\left[4x-4+2x\right]=182\Leftrightarrow5x-3\left(6x-4\right)=182\)\(\Leftrightarrow5x-18x+12-182=0\)
\(\Leftrightarrow-13x=170\Rightarrow x=\dfrac{-170}{13}\)
1) 3x - 6 = 5x + 2
=> 3x - 5x = 2 + 6
=> -2x = 8
=> x = -4
2) 15 - x = 4x - 5
=> 15 + 5 = 4x + x
=> 20 = 5x
=> x = 4
3) x - 15 = 6 + 4x
=> x - 4x = 6 + 15
=> -3x = 21
=> x = -7
4) -12 + x = 5x - 20
=> x - 5x = -20 + 12
=> -4x = -8
=> x = 2
5) 7x - 4 = 20 + 3x
=> 7x - 3x = 20 + 4
=> 4x = 24
=> x = 6
1) 3x- 6 = 5x + 2
5x - 3x = -6 - 2
2x = -8 => x = -4
Tương tự như trên
(5x-2)(3x+1)+(7-15x)(x+3)=-20
<=> 15x2+5x-6x-2+7x+21-15x2-45x+20=0
<=>39-39x=0
<=>39(1-x)=0
<=>1-x=0
=>x=1
(5x-2)(3x+1)+(7-15x)(x+3)=-20
=>\(15x^2-6x+5x-2+7x-15^2+21-45x=-20\)
=>\(-39x+19=-20\)
=>\(-39x=-39\)
=>\(x=1\)
vậy x=1