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\(\frac{4x-24}{5x-5}\div\frac{x^2-36}{x^2-2x+1}=\frac{4\left(x-6\right)}{5\left(x-1\right)}\div\frac{\left(x-6\right)\left(x+6\right)}{\left(x-1\right)^2}\)

\(=\frac{4\left(x-6\right)}{5\left(x-1\right)}.\frac{\left(x-1\right)^2}{\left(x-6\right)\left(x+6\right)}=\frac{4\left(x-1\right)}{5\left(x+6\right)}=\frac{4x-4}{5x+30}\)

1: \(=6x^2+2x-15x-5-x^2+6x-9+4x^2+20x+25-27x^3-27x^2-9x-1\)

=-27x^3-18x^2+4x+10

2: =4x^2-1-6x^2-9x+4x+6-x^3+3x^2-3x+1+8x^3+36x^2+54x+27

=7x^3+37x^2+46x+33

5:

\(=25x^2-1-x^3-27-4x^2-16x-16-9x^2+24x-16+\left(2x-5\right)^3\)

\(=8x^3-60x^2+150-125+12x^2-x^3+8x-60\)

=7x^3-48x^2+8x-35

29 tháng 9 2018

a.\(\dfrac{5\left(x-3\right)}{4\left(x+1\right)}\) : \(\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x+1\right)^2}\)

= \(\dfrac{5\left(x-3\right)}{4\left(x+1\right)}\). \(\dfrac{\left(x+1\right)^2}{\left(x-3\right)\left(x+3\right)}\)

= \(\dfrac{5\left(x+1\right)}{4\left(x+3\right)}\)

b. \(\dfrac{6\left(x+8\right)}{7\left(x-1\right)}\). \(\dfrac{\left(x-1\right)^2}{\left(x-8\right)\left(x+8\right)}\)

= \(\dfrac{6\left(x-1\right)}{7\left(x-8\right)}\)

c.Tương tự hai câu trên nka!!

d. (\(\dfrac{1}{x\left(x+1\right)}\)-\(\dfrac{2-x}{x+1}\)).(\(\dfrac{x}{x-1}\))

=( \(\dfrac{1}{x\left(x+1\right)}\)-\(\dfrac{2x-x^2}{x\left(x+1\right)}\)). ....

= \(\dfrac{\left(1-x\right)^2}{x\left(x+1\right)}\). ...

= \(\dfrac{x-1}{x+1}\)

29 tháng 9 2018

Lê Cẩm TúThiên ThảoPhạm Thái DươnMai LinhgGuyoSky SơnTùngKhôi Bùi Mysterious PersonPhong ThầnPhùng Khánh Linhtran nguyen baNguyễn Xuân Sángo quanDƯƠNG PHAN KHDũng NguyễnÁNH DƯƠNGlê thị hương giang

1)\(\left(4x-10\right)\left(24+5x\right)=0\)

\(\Leftrightarrow2\left(2x-5\right)\left(24+5x\right)=0\)

Vì 2≠0

nên \(\left[{}\begin{matrix}2x-5=0\\24+5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\5x=-24\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{5}{2}\\x=\frac{-24}{5}\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{5}{2};\frac{-24}{5}\right\}\)

2) \(0,5x\left(x-3\right)=\left(x-3\right)\left(2,5x-4\right)\)

\(\Leftrightarrow0,5x\left(x-3\right)-\left(x-3\right)\left(2,5x-4\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left[0,5x-\left(2,5x-4\right)\right]=0\)

\(\Leftrightarrow\left(x-3\right)\left(0,5x-2,5x+4\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(-2x+4\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(4-2x\right)=0\)

\(\Leftrightarrow\left(x-3\right)\cdot2\cdot\left(2-x\right)=0\)

Vì 2≠0

nên \(\left[{}\begin{matrix}x-3=0\\2-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=2\end{matrix}\right.\)

Vậy: x∈{2;3}

3) \(4x^2-1=\left(2x+1\right)\left(3x-5\right)\)

\(\Leftrightarrow\left(2x+1\right)\left(2x-1\right)-\left(2x+1\right)\left(3x-5\right)=0\)

\(\Leftrightarrow\left(2x+1\right)\left[2x-1-\left(3x-5\right)\right]=0\)

\(\Leftrightarrow\left(2x+1\right)\left(2x-1-3x+5\right)=0\)

\(\Leftrightarrow\left(2x+1\right)\left(4-x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=0\\4-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-1\\x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{2}\\x=4\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{-1}{2};4\right\}\)

4) \(\left(2-3x\right)\left(x+11\right)=\left(3x-2\right)\left(2-5x\right)\)

\(\Leftrightarrow\left(2-3x\right)\left(x+11\right)-\left(3x-2\right)\left(2-5x\right)=0\)

\(\Leftrightarrow\left(2-3x\right)\left(x+11\right)+\left(2-3x\right)\left(2-5x\right)=0\)

\(\Leftrightarrow\left(2-3x\right)\left(x+11+2-5x\right)=0\)

\(\Leftrightarrow\left(2-3x\right)\left(13-4x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2-3x=0\\13-4x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=2\\4x=13\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{2}{3}\\x=\frac{13}{4}\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{2}{3};\frac{13}{4}\right\}\)

a: \(B=\left(\dfrac{x+1}{2\left(x-1\right)}+\dfrac{3}{\left(x-1\right)\left(x+1\right)}-\dfrac{x+3}{2\left(x+1\right)}\right)\cdot\dfrac{4\left(x-1\right)\left(x+1\right)}{5}\)

\(=\dfrac{x^2+2x+1+6-x^2-2x+3}{2\left(x+1\right)\left(x-1\right)}\cdot\dfrac{4\left(x-1\right)\left(x+1\right)}{5}\)

\(=\dfrac{10}{1}\cdot\dfrac{2}{5}=10\cdot\dfrac{2}{5}=4\)

b: \(\dfrac{x^2-36}{2x+10}\cdot\dfrac{3}{6-x}\)

\(=\dfrac{\left(x-6\right)\left(x+6\right)}{2\left(x+5\right)}\cdot\dfrac{-3}{x-6}\)

\(=\dfrac{-3\left(x+6\right)}{2\left(x+5\right)}\)

c: \(\dfrac{5x+10}{4x-8}\cdot\dfrac{4-2x}{x+2}\)

\(=\dfrac{5\left(x+2\right)}{4\left(x-2\right)}\cdot\dfrac{-2\left(x-2\right)}{x+2}=\dfrac{-10}{4}=\dfrac{-5}{2}\)

d: \(\dfrac{1-4x^2}{x^2+4x}:\dfrac{2-4x}{3x}\)

\(=\dfrac{1-4x^2}{x\left(x+4\right)}\cdot\dfrac{3x}{2\left(1-2x\right)}\)

\(=\dfrac{\left(1-2x\right)\left(1+2x\right)}{x+4}\cdot\dfrac{3}{2\left(1-2x\right)}=\dfrac{3\left(2x+1\right)}{x+4}\)

2: =(2x+1)^2-y^2

=(2x+1+y)(2x+1-y)

3: =x^2(x^2+2x+1)

=x^2(x+1)^2

4: =x^2+6x-x-6

=(x+6)(x-1)

5: =-6x^2+3x+4x-2

=-3x(2x-1)+2(2x-1)

=(2x-1)(-3x+2)

6: =5x(x+y)-(x+y)

=(x+y)(5x-1)

7: =2x^2+5x-2x-5

=(2x+5)(x-1)

8: =(x^2-1)*(x^2-4)

=(x-1)(x+1)(x-2)(x+2)

9: =x^2(x-5)-9(x-5)

=(x-5)(x-3)(x+3)

nhìn mà mù mắt , rắc rối vl

29 tháng 12 2019

\(\frac{x^2-36}{2x+10}.\frac{3}{6-x}\)

\(=\frac{\left(x^2-36\right).3}{\left(2x+10\right)\left(6-x\right)}\)

\(=\frac{3\left(x+6\right)\left(x-6\right)}{\left(2x+10\right)\left(6-x\right)}\)

\(=-\frac{3\left(x+6\right)\left(x-6\right)}{2\left(x+5\right)\left(x-6\right)}\)

\(=-\frac{3\left(x+6\right)}{2\left(x+5\right)}\)

17 tháng 2 2020

\(A.\left(2,3x-6,5\right)\left(0,1x+2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2,3x-6,5=0\\0,1x+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}2,3x=6,5\\0,1x=-2\end{cases}}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{6,5}{2,3}\\x=-20\end{cases}}\)

Bài 1: Giải các phương trình: a)(5x^ 2 -45).( 4x-1 5 - 2x+1 3 )=0 b) (x^ 2 -2x+6).(2x-3)=4x^ 2 -9 d) 3 5x-1 + 2 3-5x = 4 (1-5x).(5x-3) c) (2x + 19)/(5x ^ 2 - 5) - 17/(x ^ 2 - 1) = 3/(1 - x) e) 3/(2x + 1) = 6/(2x + 3) + 8/(4x ^ 2 + 8x + 3) (x^ 2 -3x+2).(x^ 2 -9x+20)=40 (2x + 5)/95 + (2x + 6)/94 + (2x + 7)/93 = (2x + 93)/7 + (2x + 94)/6 + (2x + 95)/5 Bài 2: Giải các phương trình sau: g) a) (x + 2) ^ 2 + |5 - 2x| = x(x + 5) + 5 - 2x b) (x - 1) ^ 2 + |x + 21| - x ^ 2 - 13 =...
Đọc tiếp

Bài 1: Giải các phương trình: a)(5x^ 2 -45).( 4x-1 5 - 2x+1 3 )=0 b) (x^ 2 -2x+6).(2x-3)=4x^ 2 -9 d) 3 5x-1 + 2 3-5x = 4 (1-5x).(5x-3) c) (2x + 19)/(5x ^ 2 - 5) - 17/(x ^ 2 - 1) = 3/(1 - x) e) 3/(2x + 1) = 6/(2x + 3) + 8/(4x ^ 2 + 8x + 3) (x^ 2 -3x+2).(x^ 2 -9x+20)=40 (2x + 5)/95 + (2x + 6)/94 + (2x + 7)/93 = (2x + 93)/7 + (2x + 94)/6 + (2x + 95)/5 Bài 2: Giải các phương trình sau: g) a) (x + 2) ^ 2 + |5 - 2x| = x(x + 5) + 5 - 2x b) (x - 1) ^ 2 + |x + 21| - x ^ 2 - 13 = 0 d) |3x + 2| + |1 - 2x| = 5 - |x| c) |5 - 2x| = |1 - x| Bài 3: Cho biểu thức A = ((x + 2)/(x + 3) - 5/(x ^ 2 + x - 6) + 1/(2 - x)) / ((x ^ 2 - 5x + 4)/(x ^ 2 - 4)) a) Rút gọn A. b) Tim x de A = 3/2 c) Tìm giá trị nguyên c dot u a* d hat e A có giá trị nguyên. B = ((2x)/(2x ^ 2 - 5x + 3) - 5/(2x - 3)) / (3 + 2/(1 - x)) Bài 4: Cho biểu thức a) Rút gọn B. b) Tim* d tilde e B>0 . c) Tim* d hat e B= 1 6-x^ 2 . Bài 5: Cho biểu thức H = (2/(1 + 2x) + (4x ^ 2)/(4x ^ 2 - 1) - 1/(1 - 2x)) / (1/(2x - 1) - 1/(2x + 1)) a) Rút gọn H. b) Tìm giá trị nhỏ nhất của H. c)Tim* d vec e bi vec e u thic H= 3 2

4
8 tháng 3 2022

roois vãi

8 tháng 3 2022

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