\(-\left(-a+b+c\right)+\left(b+c-1\right)=\left(b-c+6\right)-\left(7-a+b\right)+c\)

Ta có:

\(-\left(-a+b+c\right)+\left(b+c-1\right)\)

\(=a-b-c+b+c-1\)

\(=a-\left(b-b\right)-\left(c-c\right)-1\)

\(=a-0-0-1\)

\(=a-1\) (1).

\(\left(b-c+6\right)-\left(7-a+b\right)+c\)

\(=b-c+6-7+a-b+c\)

\(=\left(b-b\right)-\left(c-c\right)+a+\left(6-7\right)\)

\(=0-0+a-1\)

\(=a-1\) (2).

Từ (1) và (2) \(\Rightarrow-\left(-a+b+c\right)+\left(b+c-1\right)=\left(b-c+6\right)-\left(7-a+b\right)+c\left(đpcm\right).\)

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Câu trả lời nằm ở đó !

a)

-(-a+b+c)+(b+c-1)=a-b-c+b+c-1=a-1

(b-c+6)-(7-a+b)+c=b-c+6-7+a-b+c=a-1

suy ra -(-a+b+c)+(b+c-1)=(b-c+6)-(7-a+b)+c

b)A+B=a+b-5-b-c+1=a-c-4

C-D=(b-c-4)-(b-a)=b-c-4-b+a=a-c-4

suy ra A+B=C-D

\(\frac{a}{c}=\frac{c}{b}\Rightarrow\frac{a^2}{c^2}=\frac{c^2}{b^2}\)

áp dụng t/c dãy tỉ số bằng nhau ta có:

\(\frac{a^2}{c^2}=\frac{c^2}{b^2}=\text{​​}\frac{a^2+c^2}{c^2+b^2}=\frac{a}{c}\cdot\frac{a}{c}=\frac{a}{c}\cdot\frac{c}{b}=\frac{a}{b}\)

=> \(\frac{a}{b}=\frac{a^2+c^2}{b^2+c^2}\left(đpcm\right)\)

b) \(7^6+7^5-7^4=7^4.\left(7^2+7-1\right)=7^4.55⋮55\left(đpcm\right)\)

a) Từ \(\frac{a}{c}=\frac{c}{b}\)\(\Rightarrow\left(\frac{a}{c}\right)^2=\left(\frac{c}{b}\right)^2=\frac{a^2}{c^2}=\frac{c^2}{b^2}=\frac{a^2+c^2}{c^2+b^2}\)(1)

Ta có \(\left(\frac{a}{c}\right)^2=\frac{a}{c}.\frac{a}{c}=\frac{a}{c}.\frac{c}{b}=\frac{a}{b}\)(2)

Từ (1) và (2) \(\Rightarrow\frac{a^2+c^2}{c^2+b^2}=\frac{a}{b}=\left(\frac{a}{c}\right)^2\left(đpcm\right)\)

b) Ta có \(7^6+7^5-7^4=7^4.\left(7^2+7-1\right)=7^4.55⋮55\left(đpcm\right)\)

a: \(\dfrac{a+5}{a-5}=\dfrac{b+6}{b-6}\)

=>(a+5)(b-6)=(a-5)(b+6)

=>ab-6a+5b-30=ab+6a-5b-30

=>-6a+5b=6a-5b

=>-12a=-10b

=>6a=5b

=>\(\dfrac{a}{b}=\dfrac{5}{6}\)

b: Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

=>\(a=bk;c=dk\)

\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{b^2k^2+b^2}{d^2k^2+d^2}=\dfrac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}=\dfrac{b^2}{d^2}\)

\(\dfrac{ab}{cd}=\dfrac{bk\cdot b}{dk\cdot d}=\dfrac{b^2k}{d^2k}=\dfrac{b^2}{d^2}\)

Do đó: \(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{ab}{cd}\)

Lời giải:

$3\text{VT}=\frac{3a}{3a+1}+\frac{3b}{3b+1}+\frac{3c}{3c+1}$

$=1-\frac{1}{3a+1}+1-\frac{1}{3b+1}+1-\frac{1}{3c+1}$

$=3-\left[\frac{1}{3a+1}+\frac{1}{3b+1}+\frac{1}{3c+1}\right]$
Áp dụng BĐT Cauchy-Schwarz:

$\frac{1}{3a+1}+\frac{1}{3b+1}+\frac{1}{3c+1}\geq \frac{9}{3a+1+3b+1+3c+1}=\frac{9}{3(a+b+c)+3}=\frac{9}{3.6+3}=\frac{3}{7}$

$\Rightarrow 3\text{VT}\leq 3-\frac{3}{7}=\frac{18}{7}$

$\Rightarrow \text{VT}\leq \frac{6}{7}$ (đpcm)

Dấu "=" xảy ra khi $a=b=c=2$

\(a\)) \(Ta\) \(có\)\(:\) \(\left(a-b\right)-\left(c-d\right)=a-b-c+d\)

\(=a+\left(-b\right)+\left(-c\right)+d\)

\(=\left(a+d\right)+\left[\left(-b\right)+\left(-c\right)\right]\)

\(=\left(a+d\right)+\left[-\left(b+c\right)\right]\)

\(=\left(a+d\right)-\left(b+c\right)\)

Bạn ơi đề bài 1  và bài 2 đều thiếu rùi kìa

Bài 1 :

7^6+7^5-7^4=7^4.49+7^4.7-7^4.1
                   =7^4.(49+7-1)
                   =7^4.55
Vì 7^4.55 chia hết 5 Vậy 7^6+7^5-7^4 chia hết 5