Question

[lê duy mạnh]

GPT 

A,\(\sqrt{3X^2-2X+1}\)+4X=\(\sqrt{3X^2+2X}\)+1

B,\(\sqrt{X^2+X-2}\)+X^2=\(\sqrt{2\left(X-1\right)}\)+1

C,X+3+\(\sqrt{1-X^2}\)=3\(\sqrt{1+X}\)+\(\sqrt{1-X}\)

D,(\(\sqrt{4+X}-1\))(\(\sqrt{1-X}\)+1)=2X

AI LM NHANH VÀ ĐÚNG MK SẼ TICKC

Answer 1

[lê duy mạnh]

MN ƠI GIÚP MK NHA MAI MIK ĐI HOK R

Answer 2

[Lê Trung Hiếu]

nhìn mà nhác giải vl :v

a) \(\sqrt{3x^2-2x+1}+4x=\sqrt{3x^2+2x}+1\)

<=> \(\sqrt{3x^2-2x+1}=\sqrt{3x^2+2x}+1-4x\)

<=> \(\left(\sqrt{3x^2-2x+1}\right)^2=\left(\sqrt{3x^2+2x}+1-4x\right)^2\)

<=> \(3x^2-2x+1=19x^2-8\sqrt{3x^2+2x}.x-6x+2\sqrt{3x^2+2x}+1\)

<=> \(-16x^2+8\sqrt{3x^2+2x}.x+4x-2\sqrt{3x^2+2x}=0\)

<=> \(-2\left(4x-1\right)\left(2x-\sqrt{3x^2+2x}\right)=0\)

<=> \(\hept{\begin{cases}x=\frac{1}{4}\\x=0\\x=2\end{cases}}\) <=> \(\orbr{\begin{cases}x=\frac{1}{4}\\x=0\end{cases}}\) (vì k có ngoặc vuông 3 nên mình dùng tạm ngoặc nhọn, thông cảm)

<=> \(\orbr{\begin{cases}x=\frac{1}{4}\\x=2\end{cases}}\)

b) \(\sqrt{x^2+x-2}+x^2=\sqrt{2\left(x-1\right)}+1\)

<=> \(\sqrt{x^2+x-2}=\sqrt{2\left(x-1\right)}+1-x^2\)

<=> \(\left(\sqrt{x^2+x-2}\right)^2=\left[\sqrt{2\left(x-1\right)}+1-x^2\right]^2\)

<=> \(x^2+x-2=x^4-2\sqrt{2}.x^2.\sqrt{x-1}-2x^2+2x+2\sqrt{2}.\sqrt{x-2}-1\)

<=> \(x^4-2\sqrt{2}.x^2.\sqrt{x-1}-2x^2+2x+2\sqrt{2}.\sqrt{x-1}-1=x^2+x-2\)

<=> \(-2\sqrt{2}.x^2.\sqrt{x-1}+2\sqrt{2}.\sqrt{x-1}-1=-x^4+3x^2-x-2\)

<=> \(-2\sqrt{2}.x^2.\sqrt{x-1}+2\sqrt{2}.\sqrt{x-1}=-x^4+3x^2-x-1\)

<=> \(-2\sqrt{2}.\sqrt{x-1}.\left(x^2+1\right)=-x^4+3x^2-x-1\)

<=> \(\left[-2\sqrt{2}.\sqrt{x-1}\left(x^2+1\right)\right]^2=\left(-x^4+3x^2-x-1\right)^2\)

<=> \(8x^5-8x^4-16x^3+16x^2+8x-8=x^8-6x^6+2x^5+11x^4-6x^3-5x^2+2x+1\)

<=> x = 1

d) mình làm tắt cho nhanh 

d) \(\left(\sqrt{4+x}-1\right)\left(\sqrt{1-x}+1\right)=2x\)

<=> \(\sqrt{4+x}.\sqrt{x-1}+\sqrt{4+x}-\sqrt{x-1}-1=2x\)

<=> \(\sqrt{4+x}.\sqrt{1-x}+\sqrt{4+x}-\sqrt{1-x}=2x+1\)

<=> \(\sqrt{4+x}.\sqrt{x-1}+\sqrt{4+x}=2x+1+\sqrt{x-1}\)

<=> \(\left(\sqrt{4+x}.\sqrt{1-x}+\sqrt{4+x}\right)^2=\left(2x+1+\sqrt{1-x}\right)^2\)

<=> \(2\sqrt{-x+1}.\left(x+4\right)=5x^2+4x\sqrt{-x+1}+5x+2\sqrt{-x+1}-6\)

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

<=> \(\sqrt{-x+1}=\frac{5x^2+4x\sqrt{-x+1}+5x+2\sqrt{-x+1}-6}{2\left(4+x\right)}\)

<=> \(2\sqrt{-x+1}.\left(4+x\right)=5x^2+4x\sqrt{-x+1}+5x+2\sqrt{-x+1}-6\)

<=> \(-2x\sqrt{-x+1}+6\sqrt{-x+1}=5x^2+5x-6\)

<=> \(\frac{2\sqrt{-x+1}.\left(-x+3\right)}{2\left(-x+3\right)}=\frac{5x^2}{2\left(-x+3\right)}+\frac{5x}{2\left(-x+3\right)}-\frac{6}{2\left(-x+3\right)}\)

<=> \(\sqrt{-x+1}=\frac{5x^2+5x-6}{2\left(x-3\right)}\)

<=> \(\left(\sqrt{-x+1}\right)^2=\left[\frac{5x^2+5x-6}{2\left(3-x\right)}\right]^2\)

<=> \(-x+1=\frac{25x^4+50x^3-35x^2-60x+36}{36-24+4x}\)

<=> \(\hept{\begin{cases}x=0\\x=\frac{21}{25}\\x=-3\end{cases}}\)=> x = 21/25 (lý do dùng ngoặc nhọn như lý do mình ghi ở trên =))) )

=> x = 21/25