Tìm x
3x^3-7x^2+17x-5=0
Tìm x
/9-7x/=5x-38x-/4x+1/=x+2/17x-5/-/17x+5/=0/3x+4/=2/2x-9/1.\(\left|9-7x\right|=5x-3\)
\(\Leftrightarrow\orbr{\begin{cases}9-7x=5x-3\\9-7x=-5x-3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}-7x-5x=-9-3\\-7x+5x=-9-3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}-12x=-12\\-2x=-12\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-12:\left(-12\right)\\x=-12:\left(-2\right)\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=6\end{cases}}\)
2.\(8x-\left|4x+1\right|=x+2\)
\(\Rightarrow\left|4x+1\right|=8x-x+2\)
\(\Rightarrow\left|4x+1\right|=7x+2\)
\(\Leftrightarrow\orbr{\begin{cases}4x+1=7x+2\\4x+1=-7x+2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}4x-7x=2-1\\4x+7x=2-1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}-3x=1\\11x=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=1:\left(-3\right)\\x=1:11\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{3}\\x=\frac{1}{11}\end{cases}}\)
Tìm x:
a. |x - 1| = 2x - 5
b. ||x + 5| - 4| = 3
c. |9 - 7x| = 5x - 3
d. 8x - |4x + 1| = x + 2
e. |17x - 5| - |17x +5| = 0
f. |3x + 4| = 2.|2x - 9|
3x3 -7x2 +17x - 5=0
tim x
|9-7X|=5x-3
|17x-5|-|17x+5|=0
|3x+4|=2.|2x-9|
5^x+2=625
(x-1)^x+2=(x-1)6x+4
a: =>|7x-9|=5x-3
\(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{3}{5}\\\left(7x-9-5x+3\right)\left(7x-9+5x-3\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{3}{5}\\\left(2x-6\right)\left(12x-12\right)=0\end{matrix}\right.\Leftrightarrow x\in\left\{3;1\right\}\)
b: =>|17x-5|=|17x+5|
=>17x-5=17x+5(vô lý) hoặc 17x-5=-17x-5
=>34x=0
hay x=0
c: =>|3x+4|=|4x-18|
=>4x-18=3x+4 hoặc 4x-18=-3x-4
=>x=22 hoặc 7x=14
=>x=22 hoặc x=2
Tìm x : a) |9-7x| = 5x-3
b) 8x - | 4x +1 | = x+2
c) | 17x - 5 | - | 17x + 5 | = 0
d) | 3x +4 | = 2 | 2x - 9 |
e) | 10x + 7 | < 37
f) |3-8x| \(\le\)19
g) | x + 3 | - 2x = | x-4 |
chào các bạn,có 2 tốt bụng thì tk mik nhé,cần lắm những người như thế
a) |9-7x| = 5x-3
* 9-7x=5x-3 <=>x=1
* 9-7x=-(5x-3) <=> x=3
Giúp mình nha! Mình đang cần gấp đó! mặc dù hơi nhiều nhưng mong các bạn giúp mình.
Tìm x
*Chú ý :/ / là dấu giá trị tuyệt đối
/9-7x/=5x-38x-/4x+1/=x+2/17x-5/-/17x+5/=0/3x+4/=2/2x-9/tìm x biết |x-1 =2x-5 b) ||x+5|-4| =3 c) |9-7x| =5x-3 d) 8x - |4x +1| =x+2 e) |17x - 5| - |17x + 5| =0 g) |3x +4| =2 |2x-9| h) Tìm x biết |10x+7|< 37 i) |3-8x|<_ ( bé thua hoặc bằng) 19 k) |x +3| -2x =|x-4|
Tim x biet
a) 28x^3+15x^2+75x+125=0
b)4x^2-x-5=0
Phan tich da thuc thanh nhan tu
a) x^3+5x^2+3x-9
b)x^3-7x-6
c)3x^3-7x^2+17x-5
\(b,4x^2-x-5=0\)
\(\Leftrightarrow4x^2-5x+4x-5=0\)
\(\Leftrightarrow x\left(4x-5\right)+4x-5=0\)
\(\Leftrightarrow\left(4x-5\right)\left(x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=-1\\x=\frac{5}{4}\end{cases}}\)
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,\(3x^3-7x^2+17x-5\)
\(\Leftrightarrow3x^3-x^2-6x^2+2x+15x-5\)
\(\Leftrightarrow x^2\left(3x-1\right)-2x\left(3x-1\right)+5\left(3x-1\right)\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2-2x+5\right)\)
\(4x^2-x-5=0\)
<=> \(4x^2+4x-5x-5=0\)
<=> \(4x\left(x+1\right)-5\left(x+1\right)=0\)
<=> \(\left(x+1\right)\left(4x-5\right)=0\)
tự lm tiếp
Tim x biet
a) 28x^3+15x^2+75x+125=0
b)4x^2-x-5=0
Phan tich da thuc thanh nhan tu
a) x^3+5x^2+3x-9
b)x^3-7x-6
c)3x^3-7x^2+17x-5
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)\(3x^3-7x^2+17x-5\)
\(=\left(3x-1\right)\left(x^2-2x+5\right)\)
\(4x^2-x-5=0\)
<=> \(4x^2+4x-5x-5=0\)
<=> \(4x\left(x+1\right)-5\left(x+1\right)=0\)
<=> \(\left(x+1\right)\left(4x-5\right)=0\)
tự giải nốt