Bài giải

Thay \(x=\frac{a}{m}\text{ ; }y=\frac{b}{m}\text{ ; }z=\frac{a+b}{m}\) vào  \(P\) ta được : 

\(P=\frac{\frac{a}{m}+\frac{b}{m}}{\frac{b}{m}+\frac{a+b}{m}}=\frac{\frac{a+m}{m}}{\frac{a+2b}{m}}=\frac{a+b}{m}\cdot\frac{m}{a+2b}=\frac{a+b}{a+2b}\)

Áp dụng : 

\(\frac{\frac{1}{4}+\frac{1}{2}}{\frac{1}{2}+\frac{3}{4}}=\frac{\frac{3}{4}}{\frac{5}{4}}=\frac{3}{4}\cdot\frac{4}{5}=\frac{3}{5}\)

Cảm ơn bạn!

Ai giúp mình hai câu cuối với!

\(\frac{x^2+xy+y^2}{x^2-xy}\)

x - 2y = 0 <=> x = 2y

Thế vào ta được :

\(\frac{x^2+xy+y^2}{x^2-xy}=\frac{\left(2y\right)^2+2y\cdot y+y^2}{\left(2y\right)^2-2y\cdot y}=\frac{4y^2+2y^2+y^2}{4y^2-2y^2}=\frac{7y^2}{2y^2}=\frac{7}{2}\)

Vậy giá trị của biểu thức = 7/2 khi x - 2y = 0 

thank u 

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Xét số nguyên dương thỏa mãn điều kiện \(1\le k< n-1\)

\(\Leftrightarrow n-k-1>0\Leftrightarrow nk-k^2-k>0\Leftrightarrow nk-k^2+n-k-n>0\)

                                     \(\Leftrightarrow k\left(n-k\right)+n-k>n\Leftrightarrow\left(k+1\right)\left(n-k\right)>n\)

Lần lượt cho k = 1, 2, 3, ..., ( n - 2 ):

Với n > 2, ta có: \(2\left(n-1\right)>n\)

                           \(3\left(n-2\right)>n\)

                           \(4\left(n-3\right)>n\)

                              \(................\)

\(\left(n-1\right)\left[n-\left(n-2\right)\right]>n\)

\(\Leftrightarrow2.3.4...\left(n-1\right).2.3.4...\left(n-1\right)>n^{n-2}\)

\(\Leftrightarrow\left[2.3.4...\left(n-1\right)\right]^2>n^{n-2}\)

\(\Leftrightarrow\left[\left(n-1\right)!\right]^2>n^{n-2}\)

Nhân 2 vế với \(n^2\), ta có: \(\left(n!\right)^2>n^2\left(đpcm\right)\)

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Bài 1 : https://h.vn/hoi-dap/question/576866.html

Bài 2 : https://h.vn/hoi-dap/question/781198.html

Tham khảo nhé .Đang bận ko làm đc

                                                                    Bài giải

x O y M A C B D

\(B=7-6x-x^2\)

\(B=-\left(x^2+6x-7\right)\)

\(B=-\left(x^2+6x+9-16\right)\)

\(B=-\left(x+3\right)^2+16\le16\)

Max B = 16 \(\Leftrightarrow x=-3\)

B = 7 - 6x - x²

= -( x² + 6x + 9 ) + 16

= -( x + 3 )² + 16

-( x + 3 )² ≤ 0 ∀ x => -( x + 3 )² + 16 ≤ 16

Đẳng thức xảy ra <=> x + 3 = 0 => x = -3

=> MaxB = 16 <=> x = -3

A = 4x² + 4x + 11

= 4( x² + x + 1/4 ) + 10

= 4( x + 1/2 )² + 10

4( x + 1/2 )² ≥ 0 ∀ x => 4( x + 1/2 )² + 10 ≥ 10

Đẳng thức xảy ra <=> x + 1/2 = 0 => x = -1/2

=> MinA = 10 <=> x = -1/2

\(A=4x^2+4x+11=4x^2+4x+1+10=\left(2x+1\right)^2+10\)

Vì \(\left(2x+1\right)^2\ge0\forall x\)\(\Rightarrow\left(2x+1\right)^2+10\ge10\forall x\)

Dấu " = " xảy ra \(\Leftrightarrow2x+1=0\)\(\Leftrightarrow2x=-1\)\(\Leftrightarrow x=\frac{-1}{2}\)

Vậy \(minA=10\)\(\Leftrightarrow x=\frac{-1}{2}\)

                                                                 Bài giải

a b A B 1 3 2 4

a, Nếu \(\widehat{A_1}=120^o\text{ ; }\widehat{B_3}=130^o\text{ }\text{thì }a\text{ không song song }b\)

Muốn \(a\text{ }//\text{ }b\text{ thì }\orbr{\begin{cases}\widehat{A_1}=\widehat{B_3}=130^o\\\widehat{A_1}=\widehat{B_3}=120^o\end{cases}}\) để hai góc bằng nhau ( so le ngoài ) 

b, Nếu \(\widehat{A_2}=65^o\text{ ; }\widehat{B_3}=64^o\) thì a không song song b

Muốn \(a\text{ }//\text{ }b\text{ thì }\orbr{\begin{cases}\widehat{A}_2=\widehat{B_3}=65^o\\\widehat{A_2}=\widehat{B_3}=64^o\end{cases}}\) để hai góc bằng nhau