1. Ta có: \(\left(a-b\right)^2=\left(a+b\right)^2-4ab\)

Theo đề ta có: \(\left(a-b\right)^2=\left(a+b\right)^2-4ab=5^2-4.2=17\)

Vậy \(\left(a-b\right)^2=17\)

2. Ta có: \(\left(a+b\right)^2=\left(a-b\right)^2+4ab\)

Theo đề ta có: \(\left(a+b\right)^2=\left(a-b\right)^2+4ab=6^2+4.16=100\)

\(\Rightarrow\left[{}\begin{matrix}a+b=10\\a+b=-10\end{matrix}\right.\)

Vậy \(a+b=10\) hoặc \(a+b=-10\)

3. \(a^2+b^2+1=ab+a+b\)

\(\Rightarrow2\left(a^2+b^2+1\right)=2\left(ab+a+b\right)\)

\(\Rightarrow2a^2+2b^2+2=2ab+2a+2b\)

\(\Rightarrow2a^2+2b^2+2-2ab-2a-2b=0\)

\(\Rightarrow\left(a^2-2ab+b^2\right)+\left(a^2-2a+1\right)+\left(b^2-2b+1\right)=0\)

\(\Rightarrow\left(a-b\right)^2+\left(a-1\right)^2+\left(b-1\right)^2=0\) (1)

Vì \(\left(a-b\right)^2\ge0\) \(\forall a;b\)
\(\left(a-1\right)^2\ge0\) \(\forall a\)

\(\left(b-1\right)^2\ge0\) \(\forall b\)

\(\Rightarrow\left(a-b\right)^2+\left(a-1\right)^2+\left(b-1\right)^2\ge0\) \(\forall a;b\) (2)

Từ (1)(2) \(\Rightarrow\left\{{}\begin{matrix}\left(a-b\right)^2=0\\\left(a-1\right)^2=0\\\left(b-1\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a-b=0\\a-1=0\\b-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=b\\a=1\\b=1\end{matrix}\right.\Rightarrow a=b=1\)

Vậy.... đpcm

Chúc bạn học tốt ahihi

a) \(9x^2+6x+1=\left(3x\right)^2+2.3x.1+1^2=\left(3x+1\right)^2\)

b) \(x^2-x+\dfrac{1}{4}=x^2-2.\dfrac{1}{2}x+\left(\dfrac{1}{2}\right)^2=\left(x-0,5\right)^2\)

c) \(x^2y^4-2xy^2+1=\left(xy^2\right)^2-2.xy^2.1+1^2=\left(xy^2-1\right)^2\)

d) \(x^2+\dfrac{2}{3}x+\dfrac{1}{9}=x^2+2.x.\dfrac{1}{3}+\left(\dfrac{1}{3}\right)^2=\left(x+\dfrac{1}{3}\right)^2\)

\(\sqrt{x+27}=3\sqrt{3}\) (ĐKXĐ: \(x\ge-27\))

\(\Rightarrow\left(\sqrt{x+27}\right)^2=\left(3\sqrt{3}\right)^2\)

\(\Rightarrow x+27=9\times3=27\)

\(\Rightarrow x=0\) (TM ĐKXĐ)

Giới hạn câu từ?

1.\(\dfrac{4^59^4-6^92}{2^{10}3^8+6^820}=\dfrac{\left(2^2\right)^5\left(3^2\right)^4-2^93^92}{2^{10}3^8+2^83^82^25}=\dfrac{2^{10}3^8-2^{10}3^9}{2^{10}3^8+2^{10}3^85}=\dfrac{2^{10}3^8\left(1-3\right)}{2^{10}3^8\left(1+5\right)}=\dfrac{-2}{6}=\dfrac{-1}{3}\)

2.

\(5^x+5^{x+2}=650\)

\(5^x+5^x5^2=650\)

\(5^x\left(1+25\right)=650\)

\(5^x=\dfrac{650}{1+25}=25=5^2\)

\(x=2\)

3.

\(5^5-5^4+5^3=5^3\left(5^2-5+1\right)=5^321=5^3.3.7⋮7\)

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

\(2x-3-2x-5=0\)

\(0-8=0\)

Đa thức vô \(n_o\)
Thân mến chúc bn học tốt

a)

20019 ; 19830

18535 ; 92196

70000 ; 18555

72325 ; 64129

74920 ; 82139

39819 ; 64456

70393 ; 59919

b)

20020 ; 19830

18540 ; 92200

70000 ; 18550

72320 ; 64130

74920 ; 82140

39820 ; 64460

70390 ; 59920

c)

20000 ; 19800

18500 ; 92200

70000 ; 18600

72300 ; 64100

74900 ; 82100

39800 ; 64500

70400 ; 59900

d)

20000 ; 20000

19000 ; 92000

70000 ; 19000

72000 ; 64000

75000 ; 82000

40000 ; 64000

70000 ; 60000

Chúc bạn học tốt