1)9x2-6x+1=0
<=>(3x-1)2=0
<=>3x-1=0
<=>x=\(\dfrac{1}{3}\)
2)bài này dùng cosi cũng được nhưng mình làm thông thường thôi
Từ đề bài =>(a+b+c)2-3(a2+b2+c2)=0
<=>a2+b2+c2+2ab+2bc+2ca-3a2-3b2-3c2=0
<=>-a2+2ab-b2-b2+2bc-c2-c2+2ca-a2=0
<=>-(a-b)2-(b-c)2-(c-a)2=0
<=>(a-b)2+(b-c)2+(c-a)2=0(1)
Do \(\left\{{}\begin{matrix}\left(a-b\right)^2\ge0\\\left(b-c\right)^2\ge0\\\left(c-a\right)^2\ge0\end{matrix}\right.\)với mọi a,b,c
=>(1) tương đương \(\left\{{}\begin{matrix}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{matrix}\right.\)<=>a=b=c
3)a)a+b=25 ab=136
=>(a+b)2-2ab=252-2.136
=>a2+2ab+b2-2ab=625-272
=>a2+b2=353
b)a2+b2=353
=>(a2+b2)2-2a2b2=3532-2.1362
=>a4+2a2b2+b4-2a2b2=87617
=>a4+b4=87617
1 ) \(9x^2-6x+1=0\)
\(\Leftrightarrow\left(3x-1\right)^2=0\)
\(\Leftrightarrow3x-1=0\)
\(\Leftrightarrow x=\dfrac{1}{3}\)
Vậy phương trình có nghiệm \(x=\dfrac{1}{3}\) .
2) (a+b+c)2=3(a2+b2+c2)
\(\Leftrightarrow\) a2+b2+c2+2ab+2bc+2bc-3a2-3b2-3c2=0
\(\Leftrightarrow\)-(2a2+2b2+2c2-2ab-2bc-2bc)=0
\(\Leftrightarrow\)2a2+2b2+2c2-2ab-2bc-2bc=0
\(\Leftrightarrow\)(a2-2ab+b2)+(b2-2bc+c2)+(a2-2ac+c2)=0
\(\Leftrightarrow\)(a-b)2+(b-c)2+(c-b)2=0
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}a-b=0\\b-c=0\\a-c=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}a=b\\b=c\\a=c\end{matrix}\right.\)\(\Leftrightarrow\)a=b=c
3) Với a+b=25 và ab=136
Ta có a+b=25 \(\Leftrightarrow\) (a+b)2=625
\(\Leftrightarrow\)a2+2ab+b2=625
\(\Leftrightarrow\)a2+272+b2=625
\(\Leftrightarrow\)a2+b2=353
2) Chứng minh tương đương
Ta có: \(\left(a+b+c\right)^2=3\left(a^2+b^2+c^2\right)\)
\(\Leftrightarrow\) \(a^2+b^2+c^2+2ac+2bc+2ab=3a^2+3b^2+3c^2\)
\(\Leftrightarrow\) \(2ab+2ac+2bc=2a^2+2b^2+2c^2\)
\(\Leftrightarrow\) \(\left(a^2-2ab+b^2\right)+\left(a^2-2ac+c^2\right)+\left(b^2-2bc+c^2\right)=0\)
\(\Leftrightarrow\) \(\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2=0\)
Ta có: (a-b)2 \(\geq\) 0; (a-c)2 \(\geq\) 0; (b-c)2 \(\geq\) 0
\(\Rightarrow\) \(\begin{cases} (a-b)^{2}=0\\ (b-c)^{2}=0\\ (a-c)^{2}=0 \end{cases}\) \(\Leftrightarrow\) \(\begin{cases} a-b=0\\ b-c=0\\ a-c=0 \end{cases}\) \(\Leftrightarrow\) \(\begin{cases} a=b\\ b=c\\ a=c \end{cases}\) \(\Leftrightarrow\) \(a=b=c\)(đpcm)
3)
b)Với a+b=25 và ab=136
ab=136\(\Rightarrow\)(ab)2=18496
\(\Leftrightarrow\)a2b2=18496
theo ý a) ta có
a2+b2=353 \(\Rightarrow\)(a2+b2)2=124609
\(\Leftrightarrow\)a4+2a2b2+b4=124609
\(\Leftrightarrow\)a4+2*18496+b2=124609
\(\Leftrightarrow\)a4+36992+b4=124609
\(\Leftrightarrow\)a4+b4=87617
