1. \(\dfrac{x+1}{x-1}+\dfrac{3x}{x+1}=4\)

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

\(\cdotĐKXĐ:x-1\ne0\Leftrightarrow x\ne1 \)
\(x+1\ne0\Leftrightarrow x\ne-1\)

pt: x² + x + x + 1 +3x² - 3x = 4x² + 4x - 4x -4

\(\Leftrightarrow\) x² + 3x² - 4x² + x + x - 3x + 4x - 4x = -4 -1

\(\Leftrightarrow\) - 1x = - 5

\(\Leftrightarrow\) x = \(\dfrac{-5}{-1}\)

\(\Leftrightarrow\) x = 5 ( nhận )

Vậy pt có tập nghiệm S= \(\left\{5\right\}\)

2. \(\left|x+2\right|< 2x+10\)

Vì x + 2 < 2x + 10⁽¹⁾ nên x + 2 > 0

-(x + 2) < 2x + 10⁽²⁾ nên - (x + 2) <0

pt(1): x + 2 < 2x + 10

\(\Leftrightarrow\) x - 2x < 10 -2

\(\Leftrightarrow\) -x < 8

\(\Leftrightarrow\) x > -8 ( nhận )

pt(2): -(x + 2) < 2x + 10

\(\Leftrightarrow\) - x - 2 < 2x + 10

\(\Leftrightarrow\) - x - 2x < 10 + 2

\(\Leftrightarrow\) -3x < 12

\(\Leftrightarrow\) x < \(\dfrac{12}{-3}\)

\(\Leftrightarrow\) x < -4 ( nhận)

Vậy bpt có tập nghiệm S= \(\left\{x\left|x< -4\right|\right\}\)

\(\left\{x\left|x>-8\right|\right\}\)

Bài 1.

\(\dfrac{x+1}{x-1}+\dfrac{3x}{x+1}=4\)(đkxđ: x\(\ne\)\(\pm\)

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

\(\Leftrightarrow\) x² + 2x + 1 + 3x² - 3x = 4(x² - 1)

\(\Leftrightarrow\) 4x² - x + 1 = 4x² - 4

\(\Leftrightarrow\) 4x² - 4x² - x = -1 - 4

\(\Leftrightarrow\) -x = -5

\(\Leftrightarrow\) x = 5 (tmđk)

Vậy................

Bài 2.

\(\left|x+2\right|< 2x+10\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}-2x-10< x+2\\x+2>2x+10\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}-2x-x< 10+2\\x-2x>10-2\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}-3x< 12\\-x>8\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x>4\\x< 8\end{matrix}\right.\)

\(\Leftrightarrow\) 4 < x < 8

Vậy........................

ta có : x^5+2x^4+3x^3+3x^2+2x+1=0

\(\Leftrightarrow\)x^5+x^4+x^4+x^3+2x^3+2x^2+x^2+x+x+1=0

\(\Leftrightarrow\)(x^5+x^4)+(x^4+x^3)+(2x^3+2x^2)+(x^2+x)+(x+1)=0

\(\Leftrightarrow\)x^4(x+1)+x^3(x+1)+2x^2(x+1)+x(x+1)+(x+1)=0

\(\Leftrightarrow\)(x+1)(x^4+x^3+2x^2+x+1)=0

\(\Leftrightarrow\)(x+1)(x^4+x^3+x^2+x^2+x+1)=0

\(\Leftrightarrow\)(x+1)[x^2(x^2+x+1)+(x^2+x+1)]=0

\(\Leftrightarrow\)(x+1)(x^2+x+1)(x^2+1)=0

VÌ x^2+x+1=(x+\(\dfrac{1}{2}\))^2+\(\dfrac{3}{4}\)\(\ne0\) và x^2+1\(\ne0\)

\(\Rightarrow\)x+1=0

\(\Rightarrow\)x=-1

CÒN CÂU B TỰ LÀM (02042006)

b: x^4+3x^3-2x^2+x-3=0

=>x^4-x^3+4x^3-4x^2+2x^2-2x+3x-3=0

=>(x-1)(x^3+4x^2+2x+3)=0

=>x-1=0

=>x=1

Câu b:

\(a,\left(6-9x\right)^2=\left(5x-7\right)^2\Leftrightarrow\left|6-9x\right|=\left|5x-7\right|\\ \Leftrightarrow\left[{}\begin{matrix}6-9x=5x-7\\6-9x=-\left(5x-7\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}14x=13\\4x=-1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{13}{14}\\x=-\dfrac{1}{4}\end{matrix}\right.\)

\(b,\left(1+x\right)^2=\left(x-1\right)^2\Leftrightarrow\left|1+x\right|=\left|x-1\right|\)

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

\(c,\left(3x+1\right)^2-4\left(x-3\right)^2=0\Leftrightarrow\left(3x+1\right)^2=[2\left(x-3\right)]^2\)

\(\Leftrightarrow\left|3x+1\right|=\left|2\left(x-3\right)\right|\Leftrightarrow\left[{}\begin{matrix}3x+1=2\left(x-3\right)\\3x+1=-2\left(x-3\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-7\\x=1\end{matrix}\right.\)

máy tính mik khó viết nhưng bài này có mẫu chung nên dễ làm mà

bn cứ đưa mẫu ra có x-8 chung đó

sau đó tính tiếp theo bt là ra mà

bạn ơi bạn làm chi tiết ra ik mk thư rôi nhưng không đc

bạn lấy MTC= 24(x-8) đi

rồi tính tiếp

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{2}\end{matrix}\right.\)

Vậy nghiệm của pt là \(S=\left\{3;-\dfrac{5}{2}\right\}\)

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

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

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

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

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

Vậy nghiệm của pt là \(S=\left\{\dfrac{2}{3};\dfrac{13}{4}\right\}\)

\(c,\left(2x+1\right)\left(3x-2\right)=\left(5x-8\right)\left(2x+1\right)\)

\(\Leftrightarrow\left(2x+1\right)\left(3x-2\right)-\left(5x-8\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\left(2x+1\right)\left(3x-2-5x+8\right)=0\)

\(\Leftrightarrow\left(2x+1\right)\left(-2x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=0\\-2x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=3\end{matrix}\right.\)

Vậy nghiệm của pt là \(S=\left\{-\dfrac{1}{2};3\right\}\)

\(d,\left(x-1\right)\left(2x-1\right)=x\left(1-x\right)\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy nghiệm của pt là \(S=\left\{1;\dfrac{1}{3}\right\}\)

\(e,0,5x\left(x-3\right)=\left(x-3\right)\left(1,5x-1\right)\)

\(\Leftrightarrow0,5x\left(x-3\right)-\left(x-3\right)\left(1,5x-1\right)=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\-x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

Vậy nghiệm của pt là \(S=\left\{1;3\right\}\)

\(f,\left(x+2\right)\left(3-4x\right)=x^2+4x=4\)

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

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\-5x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{5}\end{matrix}\right.\)

Vậy nghiệm của pt là \(S=\left\{-2;\dfrac{1}{5}\right\}\)

\(g,\left(2x^2+1\right)\left(4x-3\right)=\left(x-12\right)\left(2x^2+1\right)\)

\(\Leftrightarrow\left(2x^2+1\right)\left(4x-3\right)-\left(x-12\right)\left(2x^2+1\right)=0\)

\(\Leftrightarrow\left(2x^2+1\right)\left(4x-3-x+12\right)=0\)

\(\Leftrightarrow\left(2x^2+1\right)\left(3x+9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x^2+1>0\forall x\\3x+9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x^2+1>0\\x=-3\end{matrix}\right.\)

Vậy nghiệm của pt là \(S=\left\{-3\right\}\)

\(h,2x\left(x-1\right)=x^2-1\)

\(\Leftrightarrow2x\left(x-1\right)-\left(x-1\right)\left(x+1\right)=0\)

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

\(\Leftrightarrow\left(x-1\right)\left(x-1\right)=0\)

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

\(\Leftrightarrow x-1=0\)

\(\Leftrightarrow x=1\)

Vậy nghiệm của pt là \(S=\left\{1\right\}\)