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31 tháng 8 2018

b) x4 - 3 = (x + 1)(x - 1)

\(\Rightarrow\) x4 - 3 = x2 - 1

\(\Rightarrow\) x4 = x2 + 2

\(\Rightarrow\) 2 = x4 - x2

\(\Rightarrow\) 2,25 = (x2 - 0,5)2

\(\Rightarrow\) \(\left[{}\begin{matrix}x^2-0,5=1,5\\x^2-0,5=-1,5\end{matrix}\right.\) mà x2 - 0,5 \(\ge\) -0,5 nên x2 - 0,5 = 1,5

\(\Rightarrow x^2=2\)

\(\Rightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\)

b: \(\Leftrightarrow\dfrac{\left(x+2\right)^2}{\left(x-1\right)\left(x+2\right)}=\dfrac{-4x^2+11x-2}{\left(x+2\right)\left(x-1\right)}\)

\(\Leftrightarrow x^2+4x+4+4x^2-11x+2=0\)

\(\Leftrightarrow5x^2-7x+6=0\)

hay \(x\in\varnothing\)

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

=>3x^2-5x+2=0

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

=>(x-1)(3x-2)=0

=>x=2/3 hoặc x=1

26 tháng 8 2018

a) ta có : \(\dfrac{x}{x-1}+\dfrac{6}{x+1}-4=0\Leftrightarrow\dfrac{x\left(x+1\right)+6\left(x-1\right)-4\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=0\)

\(\Leftrightarrow x^2+x+6x-6-4x^2+4=0\Leftrightarrow-3x^2+7x-2=0\)

ta có : \(\Delta=7^2-4\left(-3\right).\left(-2\right)=25>0\)

\(\Rightarrow\) phương trình có 2 nghiệm phân biệt

\(x=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{-7+\sqrt{25}}{-6}=\dfrac{1}{3}\) ; \(x=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-7-\sqrt{25}}{-6}=2\)

vậy \(x=\dfrac{1}{3};x=2\)

câu b bn làm tương tự nha ; chỉ cần quy đồng rồi lấy tử bằng không là đc .

a: \(\Leftrightarrow x^2+x-6+2x-6=10x-20+50\)

\(\Leftrightarrow x^2+3x-12-10x-30=0\)

\(\Leftrightarrow x^2-7x-42=0\)

\(\text{Δ}=\left(-7\right)^2-4\cdot1\cdot\left(-42\right)=217>0\)

Do đó: Phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{7-\sqrt{217}}{2}\\x_2=\dfrac{7+\sqrt{217}}{2}\end{matrix}\right.\)

b: \(\Leftrightarrow x^2-3x+5=-x^2+4\)

\(\Leftrightarrow2x^2-3x+1=0\)

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

hay \(x\in\left\{\dfrac{1}{2};1\right\}\)

12 tháng 1 2019
https://i.imgur.com/NPx7OjZ.jpg
12 tháng 1 2019
https://i.imgur.com/cKHt1qr.jpg
11 tháng 4 2022

1)

<=> \(x^2-3x=0\)

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

x= 0 

x = 3

2) <=> \(x\left(x-3\right)=4\)

=> \(x=\dfrac{4}{x}+3\)

 

11 tháng 4 2022

\(2,x^2-3x=4\)

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

\(\Delta=b^2-4ac=\left(-3\right)^2-4\left(-4\right)=25>0\)

\(\Rightarrow\)Pt có 2 nghiệm pb

\(\left\{{}\begin{matrix}x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{3+5}{2}=4\\x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-3-5}{2}=-1\end{matrix}\right.\)

Vậy \(S=\left\{4;-1\right\}\)

\(3,x^4-5x^2+6=0\)

Đặt \(t=x^2\left(t\ge0\right)\)

Pt trở thành

\(t^2-5t+6=0\)

\(\Delta=b^2-4ac=\left(-5\right)^2-4.6=1>0\)

\(\Rightarrow\)Pt ó 2 nghiệm pb

\(\left\{{}\begin{matrix}x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{5+1}{2}=3\\x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-5-1}{2}-3\end{matrix}\right.\)

\(\Rightarrow t=x^2\Leftrightarrow t=\pm\sqrt{3}\)

Vậy \(S=\left\{\pm\sqrt{3}\right\}\)

 

NV
1 tháng 3 2019

a/ \(\left(x+3\right)\left(3\left(x^2+1\right)^2+2\left(x+3\right)^2\right)=5\left(x^2+1\right)^3\)

\(\Leftrightarrow3\left(x+3\right)\left(x^2+1\right)^2+2\left(x+3\right)^3-5\left(x^2+1\right)^3=0\)

\(\Leftrightarrow3\left(x+3\right)\left(x^2+1\right)^2-3\left(x^2+1\right)^3+2\left(x+3\right)^3-2\left(x^2+1\right)^3=0\)

\(\Leftrightarrow3\left(x^2+1\right)^2\left(-x^2+x+2\right)+2\left(-x^2+x+2\right)\left(\left(x+3\right)^2+\left(x+3\right)\left(x^2+1\right)+\left(x^2+1\right)^2\right)=0\)

\(\Leftrightarrow\left(-x^2+x+2\right)\left[3\left(x^2+1\right)^2+2\left(x+3+\dfrac{x^2+1}{2}\right)^2+\dfrac{3\left(x^2+1\right)^2}{4}\right]=0\)

\(\Leftrightarrow-x^2+x+2=0\) (phần ngoặc phía sau luôn dương)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)

NV
1 tháng 3 2019

b/ \(3\left(x^2+2x-1\right)^2-2\left(x^2+3x-1\right)^2+5\left(x^2+3x-1-\left(x^2+2x-1\right)\right)^2=0\)

Đặt \(\left\{{}\begin{matrix}a=x^2+2x-1\\b=x^2+3x-1\end{matrix}\right.\)

\(3a^2-2b^2+5\left(b-a\right)^2=0\Leftrightarrow8a^2+3b^2-10ab=0\)

\(\Leftrightarrow\left(4a-3b\right)\left(2a-b\right)=0\Leftrightarrow\left[{}\begin{matrix}4a=3b\\2a=b\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4\left(x^2+2x-1\right)=3\left(x^2+3x-1\right)\\2\left(x^2+2x-1\right)=x^2+3x-1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-x-1=0\\x^2+x-1=0\end{matrix}\right.\)

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

a) Ta có: \(A=\left(\dfrac{1}{\sqrt{a}+2}+\dfrac{1}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}}{a-4}\)

\(=\dfrac{\sqrt{a}-2+\sqrt{a}+2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\cdot\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\sqrt{a}}\)

=2

b) Ta có: \(B=\left(\dfrac{4x}{\sqrt{x}-1}-\dfrac{\sqrt{x}-2}{x-3\sqrt{x}+2}\right)\cdot\dfrac{\sqrt{x}-1}{x^2}\)

\(=\dfrac{4x-1}{\sqrt{x}-1}\cdot\dfrac{\sqrt{x}-1}{x^2}\)

\(=\dfrac{4x-1}{x^2}\)

a: Ta có: \(\sqrt{4x+20}-3\sqrt{x+5}+\dfrac{4}{3}\sqrt{9x+45}=6\)

\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)

\(\Leftrightarrow3\sqrt{x+5}=6\)

\(\Leftrightarrow x+5=4\)

hay x=-1

b: Ta có: \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)

\(\Leftrightarrow\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)

\(\Leftrightarrow\sqrt{x-1}=17\)

\(\Leftrightarrow x-1=289\)

hay x=290