3x(25+15)-35(5x+3)
giup minh vs
GPT sau:
a) ( x-1)(5x+3)= (3x - 8 )(x-1)
b) 3x ( 25x + 15 )- 35 ( 5x+3) = 0
c) (2-3x ) ( x-11)=(3x-2)(2- 5x)
Giups mk vs thank cacs bn
b) PT \(\Leftrightarrow15x\left(5x+3\right)-35\left(5x+3\right)=0\)
\(\Leftrightarrow\left(15x-35\right)\left(5x+3\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=-\dfrac{3}{5}\end{matrix}\right.\)
Vậy \(S=\left\{-\dfrac{3}{5};\dfrac{7}{3}\right\}\)
c) PT \(\Leftrightarrow\left(2-3x\right)\left(x-11\right)+\left(2-3x\right)\left(2-5x\right)=0\)
\(\Leftrightarrow\left(2-3x\right)\left(-9-4x\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{9}{4}\end{matrix}\right.\)
Vậy \(S=\left\{\dfrac{2}{3};-\dfrac{9}{4}\right\}\)
a)(x-1)(5x+3)=(3x-8)(x-1)
\(\Leftrightarrow\)(x-1)(5x+3)-(3x-8)(x-1)=0
\(\Leftrightarrow\left(x-1\right)\left(5x-3-3x+8\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x-5\right)=0\)
\(\left[{}\begin{matrix}x-1=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{5}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{1;\dfrac{5}{2}\right\}\)
a) Ta có: \(\left(x-1\right)\left(5x+3\right)=\left(3x-8\right)\left(x-1\right)\)
\(\Leftrightarrow5x^2+3x-5x-3=3x^2-3x-8x+8\)
\(\Leftrightarrow5x^2-2x-3=3x^2-11x+8\)
\(\Leftrightarrow5x^2-2x-3-3x^2+11x-8=0\)
\(\Leftrightarrow2x^2+9x-11=0\)
\(\Leftrightarrow2x^2+11x-2x-11=0\)
\(\Leftrightarrow x\left(2x+11\right)-\left(2x+11\right)=0\)
\(\Leftrightarrow\left(2x+11\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+11=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-11\\x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{11}{2}\\x=1\end{matrix}\right.\)
Vậy: \(S=\left\{-\dfrac{11}{2};1\right\}\)
b) Ta có: \(3x\left(25x+15\right)-35\left(5x+3\right)=0\)
\(\Leftrightarrow3x\cdot5\cdot\left(5x+3\right)-35\left(5x+3\right)=0\)
\(\Leftrightarrow15x\left(5x+3\right)-35\left(5x+3\right)=0\)
\(\Leftrightarrow\left(5x+3\right)\left(15x-35\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x+3=0\\15x-35=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=-3\\15x=35\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{5}\\x=\dfrac{7}{3}\end{matrix}\right.\)
Vậy: \(S=\left\{-\dfrac{3}{5};\dfrac{7}{3}\right\}\)
c) Ta có: \(\left(2-3x\right)\left(x-11\right)=\left(3x-2\right)\left(2-5x\right)\)
\(\Leftrightarrow2x-22-3x^2+33x=6x-15x^2-4+10x\)
\(\Leftrightarrow-3x^2+35x-22=-15x^2+16x-4\)
\(\Leftrightarrow-3x^2+35x-22+15x^2-16x+4=0\)
\(\Leftrightarrow12x^2+19x-18=0\)
\(\Leftrightarrow12x^2+27x-8x-18=0\)
\(\Leftrightarrow3x\left(4x+9\right)-2\left(4x+9\right)=0\)
\(\Leftrightarrow\left(4x+9\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+9=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=-9\\3x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{9}{4}\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy: \(S=\left\{-\dfrac{9}{4};\dfrac{2}{3}\right\}\)
Rut gon phan thuc: (x^(3)-x^(2)-x-2)/(x^(5)-3x^(4)+4x^(3)-5x^(2)+3x-2)
Minh that su da bo tay vs bai tap nay roi! Cac ban hay giup minh nhe! Minh xin cam on!
\(\frac{x^3-x^2-x-2}{x^5-3x^4+4x^3-5x^2+3x-2}\)
\(=\frac{x^3-2x^2+x^2-2x+x-2}{x^5-2x^4-x^4+2x^3+2x^3-4x^2-x^2+2x+x-2}\)
\(=\frac{\left(x^3-2x^2\right)+\left(x^2-2x\right)+\left(x-2\right)}{\left(x^5-2x^4\right)-\left(x^4-2x^3\right)+\left(2x^3-4x^2\right)-\left(x^2-2x\right)+\left(x-2\right)}\)
\(=\frac{x^2\left(x-2\right)+x\left(x-2\right)+\left(x-2\right)}{x^4\left(x-2\right)-x^3\left(x-2\right)+2x^2\left(x-2\right)-x\left(x-2\right)+\left(x-2\right)}\)
\(=\frac{\left(x-2\right)\left(x^2+x+1\right)}{\left(x-2\right)\left(x^4-x^3+2x^2-x+1\right)}=\frac{x^2+x+1}{x^4-x^3+2x^2-x+1}\)
(1/3+1/15+1/35+1/63)*x=1 va 1/3
giup minh vs tim x nha cac ban .thanks nhieu
Cái đề bài k hỉu j luôn sao lại bằng 1 và 1/3 !!
( 1/3 + 1/15 + 1/35 + 1/63 ) * x = 1 + 1/3
( 1/1x3 + 1/3x5 + 1/5x7 + 1/7x9 ) * x = 1 +1/3
1/2 x ( 2/1x3 + 2/3x5 + 2/5x7 + 2/7x9 ) * x = 1 + 1/3
1/2 x ( 1/1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 ) * x = 1 + 1/3
1/2 x ( 1 - 1/9 ) * x = 4/3
1/2 x 8/9 * x = 4/3
4/9 * x = 4/3
x = 4/3 : 4/9
x = 9/3
x =3
Vậy x = 3
Bai 1: Giai phuong trinh
a, (x - 2)3 - (x - 3) (x2 + 3x + 9) +6 ( x + 1)2 = 15
b, (5x + 1)2 - (5x + 3) (5x - 3) = 30
c, (x + 3) (x2 - 3x + 9) - x (x - 2) (x+ 2 ) = 15
GIUP MINH VOI ! THANK YOU !^^
\(a,\left(x-2\right)^2-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=15\)\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6\left(x^2+2x+1\right)=15\)\(\Leftrightarrow-6x^2+12x+19+6x^2+12x+6=15\)
\(\Leftrightarrow24x=-10\)
\(\Leftrightarrow x=-\dfrac{5}{12}\)
Vậy:....
\(b,\left(5x+1\right)^2-\left(5x+3\right)\left(5x-3\right)=30\)
\(\Leftrightarrow25x^2+10x+1-25^2+9=30\)
\(\Leftrightarrow10x=20\)
\(\Rightarrow x=2\)
Vậy :....
\(c,\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-2\right)\left(x+2\right)=15\)\(\Leftrightarrow x^3+27-x\left(x^2-4\right)=15\)
\(\Leftrightarrow x^3+27-x^3+4x=15\)
\(\Leftrightarrow4x=15-27=-12\)
\(\Leftrightarrow x=-3\)
vậy : .....
a) (x-1)(5x+3)=(3x-8)(x-1)
b) 3x(25x+15)-35(5x+3)=0
a) (x - 1)(5x + 3) = (3x - 8)(x - 1)
\(\Leftrightarrow\left(x-1\right)\left(5x+3\right)-\left(3x-8\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(5x+3-3x+8\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x+11\right)=0\)
\(\Leftrightarrow x-1=0\Rightarrow x=1\)
và\(2x+11=0\Rightarrow x=\frac{-11}{2}\)
THUC HIEN PHEP TINH:\(\frac{^{x^2}}{5x+25}-\frac{10-2x}{x}+\frac{5x+50}{5x+x^2}\)
LAM ON GIUP MINH VS LAM HOAI MA NO CU SAI T.T
\(\frac{x^2}{5x+25}-\frac{10-2x}{x}+\frac{5x+50}{5x+x^2}=\frac{x^2}{5\left(x+5\right)}-\frac{10-2x}{x}+\frac{5x+50}{x\left(x+5\right)}\)
\(=\frac{x^3}{5x\left(x+5\right)}-\frac{5\left(x+5\right)\left(10-2x\right)}{5x\left(x+5\right)}+\frac{5\left(5x+50\right)}{5x\left(x+5\right)}\)
\(=\frac{x^3+10x^2+25x}{5x\left(x+5\right)}=\frac{x\left(x+5\right)^2}{5x\left(x+5\right)}=\frac{x+5}{5}\)
3x(25x-15)-35(5x+3)=0
Giải các phương trình sau:
a. (x−1)(5x+3)=(3x−8)(x−1)(x−1)(5x+3)=(3x−8)(x−1)
b. 3x(25x+15)−35(5x+3)=0
a. (x−1)(5x+3)=(3x−8)(x−1)(x−1)(5x+3)=(3x−8)(x−1)
⇔(x−1)(5x+3)−(3x−8)(x−1)=0⇔(x−1)[(5x+3)−(3x−8)]=0⇔(x−1)(5x+3−3x+8)=0⇔(x−1)(2x+11)=0⇔(x−1)(5x+3)−(3x−8)(x−1)=0⇔(x−1)[(5x+3)−(3x−8)]=0⇔(x−1)(5x+3−3x+8)=0⇔(x−1)(2x+11)=0
⇔x−1=0⇔x−1=0hoặc 2x+11=02x+11=0
+ x−1=0⇔x=1x−1=0⇔x=1
+ 2x+11=0⇔x=−5,52x+11=0⇔x=−5,5
Phương trình có nghiệm x = 1 hoặc x = -5,5
b. 3x(25x+15)−35(5x+3)=03x(25x+15)−35(5x+3)=0
⇔15x(5x+3)−35(5x+3)=0⇔(15x−35)(5x+3)=0⇔15x(5x+3)−35(5x+3)=0⇔(15x−35)(5x+3)=0
⇔15x−35=0⇔15x−35=0 hoặc 5x+3=05x+3=0
+ 15x−35=0⇔x=3515=7315x−35=0⇔x=3515=\(\frac{7}{3}\)
+ 5x+3=0⇔x=−355x+3=0⇔x=−\(\frac{3}{5}\)
Phương trình có nghiệm x=\(\frac{7}{3}\)x=\(\frac{7}{3}\) hoặc x=−\(\frac{3}{5}\)
ai giup minh voi
a) x(x+1)(x+2)(x+3)=24
b)5x(2-3x)=4-6x
c) (5-x)(2+3x)=4-9x2
d) 25-x2=4x(5+x)
a) bạn nhóm 2 cái cuối thành 1 nhóm, 2 cái ở giữa thành 1 nhóm, rồi đặt ẩn phụ là ra
\(x\left(x+1\right)\left(x+2\right)\left(x+3\right)=24\)
\(\Leftrightarrow\)\(\left(x^2+3x\right)\left(x^2+3x+2\right)-24=0\)
Đặt \(x^2+3x=t\) ta có:
\(t\left(t+2\right)-24=0\)
\(\Leftrightarrow\)\(t^2+2t-24=0\)
\(\Leftrightarrow\)\(\left(t-4\right)\left(t+6\right)=0\)
đến đây bn thay trở lại rồi tìm nghiệm nhé
a, x(x+3)(x+1)(x+2)-24=0
=> (x^2+3x)(x^2+3x+2)-24=0
đặt x^2+3x=a
ta có : a(a+2)-24=0
=> a^2+2a-24=0 => \(\orbr{\begin{cases}a=4\\a=-6\end{cases}}\) giải ra ta được x^2+3x=4 hay \(\orbr{\begin{cases}x=-4\\x=1\end{cases}}\)
và x^2+3x=-6 => vô nghiệm vậy x=-4 hoặc x=1
b, 5x(2-3x)-(4-6x)=0
=> 5x(2-3x)-2(2-3x)=0
=>(5x-2)(2-3x)=0
=>\(\orbr{\begin{cases}x=\frac{3}{2}\\x=\frac{2}{5}\end{cases}}\)