Đặt:\(\dfrac{a}{b}=\dfrac{c}{d}=@\Leftrightarrow\left\{{}\begin{matrix}a=b@\\c=d@\end{matrix}\right.\)

khi đó: \(\dfrac{a^{2017}+b^{2017}}{c^{2017}+d^{2017}}=\dfrac{b^{2017}@^{2017}+b^{2017}}{d^{2017}@^{2017}+d^{2017}}=\dfrac{b^{2017}\left(@^{2017}+1\right)}{d^{2017}\left(@^{2017}+1\right)}=\dfrac{b^{2017}}{d^{2017}}\)

\(\dfrac{\left(a-b\right)^{2017}}{\left(c-d\right)^{2017}}=\dfrac{\left(b@-b\right)^{2017}}{\left(d@-d\right)^{2017}}=\dfrac{\left[b\left(@-1\right)\right]^{2017}}{\left[d\left(@-1\right)\right]^{2017}}=\dfrac{b^{2017}}{d^{2017}}\)

Ta có điều phải chứng minh

\(A=1.2.3+2.3.4+...+98.99.100\)

\(4A=1.2.3.4+2.3.4.\left(5-1\right)+....+98.99.100.\left(101-97\right)\)

\(4A=1.2.3.4+2.3.4.5-1.2.3.4=...+98.99.100.101-97.98.99.100\)

\(4A=98.99.100.101\)

\(A=\dfrac{98.99.100.101}{4}=24497550\)

\(\left\{{}\begin{matrix}\widehat{A}-\widehat{B}=50^o\\\widehat{A}=\dfrac{1}{2}\widehat{C}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\widehat{B}=\widehat{A}-50^o\\\widehat{C}=2\widehat{A}\end{matrix}\right.\)

Thay vào ta có:

\(\widehat{A}+\left(\widehat{A}-50^o\right)+2\widehat{A}=180^o\)

\(\Rightarrow4\widehat{A}-50^o=180^o\)

\(\Rightarrow4\widehat{A}=230^o\Leftrightarrow\widehat{A}=\dfrac{230^o}{4}=57,5^o\)

\(\Rightarrow\left\{{}\begin{matrix}\widehat{B}=57,5^o-50^o=7,5^o\\\widehat{C}=57,5^o.2=115^o\end{matrix}\right.\)

Mk sẽ ko nghi lại phần chú thích:

\(x=y\)

\(\Leftrightarrow x^2=xy\)(Đ)

\(\Leftrightarrow x^2-y^2=xy-y^2\)(Đ)

\(\Leftrightarrow\left(x+y\right)\left(x-y\right)=y\left(x-y\right)\)(Đ)

\(\Leftrightarrow x+y=y\)(Đ)

\(\Leftrightarrow2y=y\)(Đ)

Đến đoạn này,đề bài xét thiếu,mk sẽ làm theo cách of mk

\(\Leftrightarrow2y-y=0\)

\(\Rightarrow y\left(2-1\right)=0\)

\(\Rightarrow y=0\)

nên ko phải là \(2=1\) mà là \(x=y=0\)

\(VT=-2x^2+6x-12\)

\(VT=-2\left(x^2-3x+6\right)\)

\(VT=-2\left(x^2-3x+\dfrac{9}{4}+\dfrac{15}{4}\right)\)

\(VT=-2\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{15}{4}.2\)

\(VT=-2\left(x-\dfrac{3}{2}\right)^2-\dfrac{15}{2}< 0=VP\)

Vậy phương trình vô nghiệm(đpcm)

Gọi số học sinh lớp \(7A;7B;7C\) lần lượt là \(a;b;c\)

Theo đề bài ta có:

\(\dfrac{2}{3}a=\dfrac{3}{4}b=\dfrac{4}{5}c\Leftrightarrow\dfrac{2a}{3}=\dfrac{3b}{4}=\dfrac{4c}{5}\)

Tương đương với:

\(\dfrac{2a}{3}.\dfrac{1}{12}=\dfrac{3b}{4}.\dfrac{1}{12}=\dfrac{4c}{5}.\dfrac{1}{12}\)

\(\Leftrightarrow\dfrac{2a}{36}=\dfrac{3b}{48}=\dfrac{4c}{60}\)

\(\Leftrightarrow\dfrac{a}{18}=\dfrac{b}{16}=\dfrac{c}{15}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{a}{18}=\dfrac{b}{16}=\dfrac{c}{15}=\dfrac{a+b-c}{18+16-15}=\dfrac{57}{19}=3\)

\(\Rightarrow\left\{{}\begin{matrix}a=18.3=54\\b=16.3=48\\c=15.3=45\end{matrix}\right.\)

\(d=\left(1+\dfrac{1}{3}\right)\left(1+\dfrac{1}{8}\right)\left(1+\dfrac{1}{15}\right)....\left(1+\dfrac{1}{n^2+2n}\right)\)

\(d=\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}...........\dfrac{n^2+2n+1}{n^2+2n}\)

\(d=\dfrac{2^2}{3}.\dfrac{3^2}{8}.\dfrac{4^2}{15}......\dfrac{\left(n+1\right)^2}{n\left(n+2\right)}\)

\(d=\dfrac{2^2.3^2.4^2......\left(n+1\right)^2}{3.8.15.....n\left(n+2\right)}\)

\(d=\dfrac{2.2.3.3.4.4......\left(n+1\right)\left(n+1\right)}{1.3.2.4.3.5......n\left(n+2\right)}\)

\(d=\dfrac{2.3.4......\left(n+1\right)}{1.2.3......n}.\dfrac{2.3.4.....\left(n+1\right)}{3.4.5.....\left(n+2\right)}\)

\(d=\left(n+1\right)\dfrac{2}{n+2}\)

\(d=\dfrac{2n+2}{n+2}\)

\(\left(2x+1\right)^2-\left(x-1\right)^2=\left[\left(2x+1\right)+\left(x-1\right)\right]\left[\left(2x+1\right)-\left(x-1\right)\right]\)

\(=\left(2x+1+x-1\right)\left(2x+1-x+1\right)\)

\(=3x\left(x+2\right)\)

xy + 2x + y -13 = 0

xy + 2x + y = 0 + 13

xy + 2x + y = 13

xy + y + 2x = 13

y(x + 1) = 13 - 2x

=> 13 - 2x chia hết cho x + 1.

Đến đây nếu bạn biết làm thì làm nốt, nếu chưa biết thì bảo mình mình làm nốt cho

Lời giải:

(A=-x^2+4x-5)

(A=-left(x^2-4x+5 ight))

(A=-left(x^2-4x+4+1 ight))

(A=-left(x^2-4x+4 ight)-1)

(A=-left(x-2 ight)^2-1le1)

Dấu "=" xảy ra khi: (x=2)

(B=-2x^2-6x+5)

(B=-2left(x^2+3x-dfrac{5}{2} ight))

(B=-2left(x^2+3x-dfrac{19}{4}+dfrac{9}{4} ight))

(B=-2left(x^2+3x+dfrac{9}{4} ight)+dfrac{19}{2})

(B=-2left(x+dfrac{3}{2} ight)^2+dfrac{19}{2}ledfrac{19}{2})

Dấu "=" xảy ra khi : (x=-dfrac{3}{2})