2.Ta có : \(4\sqrt{3+2\sqrt{2}}-\sqrt{56\sqrt{2}+81}\)

\(=4\sqrt{2+2\sqrt{2}+1}-\sqrt{56\sqrt{2}+81}\)

\(=4\sqrt{2}+4-\sqrt{56\sqrt{2}+81}\)

\(=4\sqrt{2}+4-\sqrt{7^2+2.4\sqrt{2}.7+\left(4\sqrt{2}\right)^2}\)

\(=4\sqrt{2}+4-7-4\sqrt{2}=4-7=-3\)

3.Ta có : \(\frac{x-49}{\sqrt{x}-7}\)

\(=\frac{\left(\sqrt{x}-7\right)\left(\sqrt{x}+7\right)}{\sqrt{x}-7}=\sqrt{x}+7\)

4.Ta có : \(\sqrt{x+2\sqrt{x+1}}\)

\(=\sqrt{x+1+2\sqrt{x+1}+1-1}\)

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

5.Ta có : \(\sqrt{x-1-2\sqrt{x-2}}\)

\(=\sqrt{x-2-2\sqrt{x-2}+1}\)

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

MN ƠI GIÚP MK NHA MAI MIK ĐI HOK R

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

a,Phương trình đã cho tương đương với:

( \(\dfrac{x+1}{58}\)+1) + ( \(\dfrac{x+2}{57}\)+1) =(\(\dfrac{x+3}{56}\)+1) +(\(\dfrac{x+4}{55}\)+1)

\(\Leftrightarrow\)(x +59) (\(\dfrac{1}{58}\)+\(\dfrac{1}{57}\)- \(\dfrac{1}{56}\)-\(\dfrac{1}{55}\))

\(\Leftrightarrow\)(x+59)=0 \(\Rightarrow\)x = -59

b,

\(\dfrac{x+a}{a-2}\)+\(\dfrac{x-a}{a+2}\)=\(\dfrac{x+2}{a+2}\)+\(\dfrac{2x-4}{a-2}\)

\(\Leftrightarrow\)(x+a)(a+2)+(x-a)(a-2) = (x+2)(a-2) +(2x-4)(a+2)

\(\Leftrightarrow\)(a+2)x = 6(a+2)

\(\Leftrightarrow\)x=6 (do a+2 \(\ne\)0)

Vậy phương trình đã cho có nghiệm duy nhất x=6 (với a\(\ne\) \(\pm\)2)

đính chinh nha bươc 2 câu a thêm =0 ở cuối

bạn giải theo delta nha :) mình vd một câu đó

\(1.x^2-11x+30=0\)

\(\Delta=\left(-11\right)^2-4.1.30=1>0\)

Do đó pt có 2 nghiệm phân biệt là:

\(x_1=\frac{11+\sqrt{1}}{2}=6;x_2=\frac{11-\sqrt{1}}{2}=5\)

cảm ơn bạn

MN ƠI GIÚP MK NHA

Bình phương 2 cái đấy ta có

x⁶ - 6x⁴y² + 9x²y⁴ = 100

y⁶ - 6x²y⁴ + 9x⁴y² = 900

Cộng vế theo vế được

x⁶ + 3x²y⁴ + 3x⁴y² + y⁶ = 1000

<=> (x² + y²)³ = 1000

<=> x² + y² = 10