Bài tập: Giải các pt sau:
1/ \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=0\)
2/ \(x\left(x+4\right)\left(x+6\right)\left(x+10\right)+128=0\)
Giải hộ mik nha! Mik đang cần gấp.
Giải các phương trình sau:
a) \(\left(4x-1\right)\left(x-3\right)=\left(x-3\right)\left(5x+2\right)\).
b)\(\left(x+3\right)\left(x-5\right)+\left(x+3\right)\left(3x-4\right)=0\)
c)\(\left(1-x\right)\left(5x+3\right)=\left(3x-7\right)\left(x-1\right)\). Giải chi tiết hộ mik nhoa, mik tik
TA CÓ:
\(a,\left(4x-1\right)\left(x-3\right)=\left(x-3\right)\left(5x+2\right)\Leftrightarrow\left(4x-1\right)\left(x-3\right)-\left(x-3\right)\left(5x+2\right)=0\)
\(\left(x-3\right)\left(4x-1-5x-2\right)=0\Leftrightarrow\left(x-3\right)\left(-x-3\right)=0\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)
\(b,\left(x+3\right)\left(x-5\right)+\left(x+3\right)\left(3x-4\right)=0\Leftrightarrow\left(x+3\right)\left(x-5+3x-4\right)=0\)
\(\left(x-3\right)\left(4x-9\right)=0\orbr{\begin{cases}x=3\\x=\frac{9}{4}\end{cases}}\)
\(c,\left(1-x\right)\left(5x+3\right)=\left(3x-7\right)\left(x-1\right)\Leftrightarrow\left(1-x\right)\left(5x+3\right)=\left(7-3x\right)\left(1-x\right)\)
\(\left(1-x\right)\left(5x+3-7+3x\right)=0\Leftrightarrow\left(1-x\right)\left(8x-4\right)=0\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}\)
Giải các phương trình sau:
a \(\left(x+2\right)\left(x+\text{4}\right)\left(x+6\right)\left(x+8\right)+16=0\)
b \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=0\)
c \(\left(4x+1\right)\left(12x-1\right)\left(3x+2\right)\left(x+1\right)-4=0\)
d \(\left(x^2-3x+2\right)\left(x^2+15x+56\right)+8=0\)
b: Ta có: \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=0\)
\(\Leftrightarrow\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24=0\)
\(\Leftrightarrow\left(x^2+7x\right)^2+22\left(x^2+7x\right)+120-24=0\)
\(\Leftrightarrow x^2+7x+6=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-6\end{matrix}\right.\)
Giải các PT sau
a)\(\left(3\left(x+1\right)-2\left(x+3\right)\right)^3+\left(2\left(x+3\right)-x+5\right)^3+\left(x-5-3\left(x+1\right)\right)^3=0\)
b)\(\left(x-2\right)^3+\left(x-4\right)^3+\left(x-7\right)^3+3\left(x-2\right)\left(x-4\right)\left(x-7\right)=0\)
Bài Tập: Giải phương trình :
a) (x + 5)(2x - 3) = 0
b) \(\left(x^2-9\right)\left(4-x\right)=0\)
c) \(\left(2x+3\right)\left(4-5x\right)=0\)
d) \(2\left(x+3\right)\left(x-4\right)=0\)
e) \(\left(x^2-9\right)\left(4-x\right)=0\)
f) \(\left(2x+3\right)\left(x^2-16\right)=0\)
a: \(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=\dfrac{3}{2}\end{matrix}\right.\)
b: \(\Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\\x=4\end{matrix}\right.\)
c: \(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\5x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{4}{5}\end{matrix}\right.\)
d: \(\Leftrightarrow\left(x+3\right)\left(x-4\right)=0\)
=>x+3=0 hoặc x-4=0
=>x=-3 hoặc x=4
e: \(\Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\\x=4\end{matrix}\right.\)
f: \(\Leftrightarrow\left(2x+3\right)\left(x-4\right)\left(x+4\right)=0\)
hay \(x\in\left\{-\dfrac{3}{2};4;-4\right\}\)
a, \(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=\dfrac{3}{2}\end{matrix}\right.\)
b, \(\Leftrightarrow\left[{}\begin{matrix}x^2-9=0\\4-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\pm3\\x=4\end{matrix}\right.\)
c, \(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\4-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{4}{5}\end{matrix}\right.\)
d, \(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=4\end{matrix}\right.\)
e, tương tự d
f, \(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\x^2-16=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\pm4\end{matrix}\right.\)
Giải các phương trình sau:
\(\left(x+6\right)\left(3x+1\right)+x+6=0\).
\(\left(x+4\right)\left(5x+9\right)-x-4=0\). Giúp mik nhoa mik cần gấp lắm
a,(x + 6)(3x +1) + x+6 = 0
(x+6)(3x +2)=0
x= -6
x= -2/3
b, x= -4 hoăc x =-8/5
a,(x+6)(3x+1)+x+6=0
=>(x+6)(3x+2)=0
=>x+6=0 hoặc 3x+2=0
=>x=-6 hoặc x=-2/3
b,(x+4)(5x+9)-x-4=0
=>(x+4)(5x+8)=0
=>x+4=0 hoặc 5x+8=0
=>x=-4 hoặc x=-8/5
Giải phương trình sau:
a) \(\left(x+2\right)\left(x+3\right)\left(x+6\right)\left(x+8\right)+16=0\)
b) \(\left(x+3\right)\left(x-1\right)\left(x+5\right)\left(x+15\right)+64x^2=0\)
c) \(\left(x-3\right)\left(x-5\right)\left(x-6\right)\left(x-10\right)-24x^2=0\)
Giúp mình với ạ. Mình đang cần gấp. Cảm ơn trước ạ :)
giải pt
\(a,\left(x^2-5x\right)^2+10\left(x^2-5x\right)+24=0\)
\(b,\left(x-4\right)\left(x-5\right)\left(x-6\right)\left(x-7\right)=1680\)
\(c,\left(x+1\right)^4+\left(x-1\right)^4=82\)
d,\(\left(4-x\right)^5+\left(x+2\right)^5=32\)
a/ Đặt (x^2 - 5x) = a thì ta có
a^2 + 10a + 24 = 0
<=> (a + 4)(a + 6) = 0
Làm nốt
b/ (x - 4)(x - 5)(x - 6)(x - 7) = 1680
<=> (x - 4)(x - 7)(x - 5)(x - 6) = 1680
<=> (x^2 - 11x + 28)(x^2 - 11x + 30) = 1680
Đặt x^2 - 11x + 28 = a thì ta có
a(a + 2) = 1680
<=> (a - 40)(a + 42) = 0
Làm nốt
c/ (x + 1)^4 + (x - 1)^4 = 82
<=> x^4 + 6x^2 - 40 = 0
Đặt x^2 = a
=> a^2 + 6a - 40 = 0
<=> (a - 4)(a + 10) = 0
Giải các phương trình sau:
1, \(\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}\)
2, \(\left(x-2\right)\left(2x-1\right)=x^2-2x\)
3, \(3x^2-4x+1=0\)
4, \(\left|2x-4\right|=0\)
5, \(\left|3x+2\right|=4\)
6, \(\left|2x-5\right|=\left|-x+2\right|\)
*Giúp mình với mình đg cần gấp ạ T_T
\(1.\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}.\Leftrightarrow\dfrac{x-1-3x}{3}=\dfrac{x-2}{2}.\Leftrightarrow\dfrac{-2x-1}{3}-\dfrac{x-2}{2}=0.\)
\(\Leftrightarrow\dfrac{-4x-2-3x+6}{6}=0.\Rightarrow-7x+4=0.\Leftrightarrow x=\dfrac{4}{7}.\)
\(2.\left(x-2\right)\left(2x-1\right)=x^2-2x.\Leftrightarrow\left(x-2\right)\left(2x-1\right)-x\left(x-2\right)=0.\)
\(\Leftrightarrow\left(x-2\right)\left(2x-1-x\right)=0.\Leftrightarrow\left(x-2\right)\left(x-1\right)=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2.\\x=1.\end{matrix}\right.\)
\(3.3x^2-4x+1=0.\Leftrightarrow\left(x-1\right)\left(x-\dfrac{1}{3}\right)=0.\Leftrightarrow\left[{}\begin{matrix}x=1.\\x=\dfrac{1}{3}.\end{matrix}\right.\)
\(4.\left|2x-4\right|=0.\Leftrightarrow2x-4=0.\Leftrightarrow x=2.\)
\(5.\left|3x+2\right|=4.\Leftrightarrow\left[{}\begin{matrix}3x+2=4.\\3x+2=-4.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}.\\x=-2.\end{matrix}\right.\)
\(1,\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}\\ \Leftrightarrow\dfrac{x-1}{3}-x=\dfrac{x-2}{2}\\ \Leftrightarrow\dfrac{2\left(x-1\right)-6x}{6}=\dfrac{3\left(x-2\right)}{6}\\ \Leftrightarrow2\left(x-1\right)-6x=3\left(x-2\right)\\ \Leftrightarrow2x-2-6x=3x-6\\ \Leftrightarrow-4x-2=3x-6\)
\(\Leftrightarrow3x-6+4x+2=0\\ \Leftrightarrow7x-4=0\\ \Leftrightarrow x=\dfrac{4}{7}\)
\(2,\left(x-2\right)\left(2x-1\right)=x^2-2x\\ \Leftrightarrow2x^2-4x-x+2=x^2-2x\\ \Leftrightarrow x^2-3x+2=0\\ \Leftrightarrow\left(x^2-2x\right)-\left(x-2\right)=0\\ \Leftrightarrow x\left(x-2\right)-\left(x-2\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
\(3,3x^2-4x+1=0\\ \Leftrightarrow\left(3x^2-3x\right)-\left(x-1\right)=0\\ \Leftrightarrow3x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(3x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\end{matrix}\right.\)
\(4,\left|2x-4\right|=0\\ \Leftrightarrow2x-4=0\\ \Leftrightarrow2x=4\\ \Leftrightarrow x=2\)
\(5,\left|3x+2\right|=4\\ \Leftrightarrow\left[{}\begin{matrix}3x+2=4\\3x+2=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=2\\3x=-6\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-2\end{matrix}\right.\)
\(6,\left|2x-5\right|=\left|-x+2\right|\\ \Leftrightarrow\left[{}\begin{matrix}2x-5=-x+2\\2x-5=x-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=7\\x=3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=3\end{matrix}\right.\)
Giải các phương trình sau:
a)\(\left(4x-1\right)\left(x-3\right)=\left(x-3\right)\left(5x+2\right)\) )
b)\(\left(x+3\right)\left(x-5\right)+\left(x+3\right)\left(3x-4\right)=0\). Giúp mik nhoa mik tích cho.
nhầm b, x= -3 hoặ x= 9/4
lúc nãy đánh nhầm