viết lại cho số thực thỏa mãn ab+bc+ca=abc và a+b+c=1

Ta có \(\left(a-1\right)\left(b-1\right)\left(c-1\right)=abc-ab-bc-ca+a+b+c-1=0\)(ĐPCM)

^.^

a, Từ ghép: mùa hè, nhà trường,khai trường,học trò, bà ngoại,ngôi trường,cổng trường,cánh cổng,thế giới,con người

b, Từ láy âm: nhẹ nhàng, rạo rực, xao xuyến, nôn nao, hồi hộp, hốt hoảng

    Từ láy vần:  bâng khuâng,chơi vơi,

    Láy toàn bộ: mãi mãi

a ) \(x^2+4x+5=x^2+2.x.2+2^2+1=\left(x+2\right)^2+1\)

\(Do\left(x+2\right)^2\ge0\Rightarrow\left(x+2\right)^2+1\ge1>0\forall x\left(đpcm\right)\)

b) \(x^2-x+1=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\)

\(Do\left(x-\frac{1}{2}\right)^2\ge0\Rightarrow\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\forall x\left(đpcm\right)\)

c)\(-\left(4x^2-12x+9\right)-1=-\left(2x-3\right)^2-1\)

\(Do-\left(2x-3\right)\le0\Rightarrow-\left(2x-3\right)-1\le-1\forall x\)

\(x^2+2.x.2+2^2+5-4\) \(\Rightarrow\left(x+2\right)^2+5-4\) \(\Rightarrow\left(x+2\right)^2+1\)

 vì \(\left(x+2\right)^2\ge0\) \(\Rightarrow\left(x+2\right)^2+1\ge1\)  \(\ge0\) \(\Rightarrow dpcm\)

b) \(x^2-2.x.\frac{1}{2}+\left(\frac{1}{2}\right)^2+1-\left(\frac{1}{2}\right)^2\) \(\Rightarrow\left(x-\frac{1}{2}\right)^2+\frac{5}{4}\)

vì \(\left(x+\frac{1}{2}\right)^2\ge0\) \(\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{5}{4}\ge\frac{5}{4}\ge0\) \(\Rightarrow dpcm\)

c) \(12x-4x^2-10=-\left(4x^2-12x+10\right)\) = \(\left[\left(2x\right)^2-2.2x.3+3^2\right]+10-3^2\)

\(\Rightarrow\left(2x-3\right)^2+10-9\) \(\Rightarrow\left(2x-3\right)^2+1\) vì \(\left(2x-3\right)^2\ge0\Rightarrow\left(2x-3\right)^2+1\ge1hay\ge0\left(1>0\right)\Rightarrow dpcm\)

a)

\(\frac{-\frac{6}{5}-\frac{6}{35}-\frac{6}{125}-\frac{6}{2009}-\frac{6}{2011}}{\frac{7}{5}+\frac{7}{35}+\frac{7}{125}+\frac{7}{2009}+\frac{7}{2011}}=\frac{-6.\left(\frac{1}{5}+\frac{1}{35}+\frac{1}{125}+\frac{1}{2009}+\frac{6}{2011}\right)}{7.\left(\frac{1}{5}+\frac{1}{35}+\frac{1}{125}+\frac{1}{2009}+\frac{1}{2011}\right)}\)\(=\frac{-6}{7}\)

đề sai nhé đề như mik ms đúng đi lm tiếp

Câu b thì sao bạn ơi??

\(n^2-5n+3=\left(4-n\right)\left(-n+1\right)-1\)

\(\left(4-1\right)\left(-n+1\right)⋮\left(4-n\right)\Rightarrow-1⋮\left(4-n\right)\)

Vậy ..

Ta có \(\left|2000x+2012\right|+\left|2013-2000x\right|\ge\left|2000x+2012+2013-2000x\right|=\left|4025\right|=4025\)

^.^

thank

1;Ta có\(5.3^x=5.3^4\)

\(\Rightarrow3^x=3^4\)

\(\Rightarrow x=4\)

2.Ta có \(9.5^x=6.5^6+3.5^6\)

\(\Rightarrow9.5^x=5^6.\left(6+3\right)\)

\(\Rightarrow9.5^x=9.5^6\)

\(\Rightarrow5^x=5^6\)\

\(\Rightarrow x=6\)

3, Ta có \(2.3^{x+2}+4.3^{x+1}=10.3^6\)

\(\Rightarrow3^{x+1}.\left(2.3+4\right)=10.3^6\)

\(\Rightarrow3^{x+1}.10=10.3^6\)

\(\Rightarrow3^{x+1}=3^6\)

\(\Rightarrow x+1=6\)

\(\Rightarrow x=5\)

a) 5.3^x = 5.3⁴

=> 3^x=3⁴

=> x=4

b) 9.5^x=6.5⁶+3.5⁶

=> 9.5^x = (6+3)5⁶

=> 9.5^x=9.5⁶

=> 5^x=5⁶

=> x=6

c) 2.3^x+2 + 4.3^x+1 = 10.3⁶

=> 2.3^x+1.3 + 4.3^x+1 = 10.3⁶ 

=> 6.3^x+1+4.3^x+1=10.3⁶

=> (6+4).3^x+1=10.3⁶

=> 10.3^x+1=10.3⁶

=> 3^x+1=3⁶

=> x+1=6

=> x=5

kí hiệu góc vuông

Kí hiệu góc vuông