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-17/10

mng ơi giúp e với ạ

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\(K=-x^2+10x-30\)

\(=-x^2+10x-25-5\)

\(=-\left(x^2-10x+25\right)-5\)

\(=-\left(x-5\right)^2-5\)

Ta thấy: \(\left(x-5\right)^2\ge0\forall x\)

\(\Rightarrow-\left(x-5\right)^2\le0\forall x\)

\(\Rightarrow-\left(x-5\right)^2-5\le-5\)

Dấu \("="\) xảy ra \(\Leftrightarrow x-5=0\Leftrightarrow x=5\)

Vậy \(K_{max}=-5\) khi \(x=5.\)

\(---\)

\(Q=-25y^2-20y-2\)

\(=-25y^2-20y-4+2\)

\(=-\left(25y^2+20y+4\right)+2\)

\(=-\left(5y+2\right)^2+2\)

Ta thấy: \(\left(5y+2\right)^2\ge0\forall y\)

\(\Rightarrow-\left(5y+2\right)^2\le0\forall y\)

\(\Rightarrow-\left(5y+2\right)^2+2\le2\forall y\)

Dấu \("="\) xảy ra \(\Leftrightarrow5y+2=0\Leftrightarrow y=-\dfrac{2}{5}\)

Vậy \(Q_{max}=2\) khi \(y=-\dfrac{2}{5}.\)

#\(Toru\)

\(K=-x^2+10x-30\\ =-\left(x^2-10x+25\right)-5\\ =-\left(x^2-2.x.5+5^2\right)-5\\ =-\left(x-5\right)^2-5\le-5\forall x\in R\\ Vậy:max_K=-5.khi.x=5\)

\(Q=-25y^2-20y-2\\ =-\left(25y^2+20y+4\right)+2\\ =-\left[\left(5y\right)^2+2.5y.2+2^2\right]+2\\ =-\left(5y+2\right)^2+2\le2\forall y\in R\\ Vậy:max_Q=2.khi.y=-\dfrac{2}{5}\)

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ghép các các lại với nhau là được

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làm rồi

lam giup minh voi

\(\frac{5}{4}-\frac{4}{5}\times\frac{4}{3}\)

\(=\frac{5}{4}-\frac{16}{15}\)

\(=\frac{11}{60}\)

\(\frac{3}{7}:\frac{1}{2}\times\frac{14}{15}\)

\(=\frac{3}{7}\times\frac{2}{1}\times\frac{14}{15}\)

\(=\frac{4}{5}\)

\(\frac{2}{5}+x=\frac{6}{7}\)

           \(x=\frac{6}{7}-\frac{2}{5}\)

           \(x=\frac{16}{35}\)

\(\frac{1}{3}:x=\frac{8}{9}\)

       \(x=\frac{1}{3}:\frac{8}{9}\)

       \(x=\frac{3}{8}\)

\(\dfrac{5}{x-3}+\dfrac{4}{x+3}=\dfrac{x-5}{x^2-9}\left(ĐKXĐ:x\ne\pm3\right)\\ \Leftrightarrow\dfrac{5\left(x+3\right)+4\left(x-3\right)}{x^2-9}=\dfrac{x-5}{x^2-9}\\ \Leftrightarrow5x+15+4x-12=x-5\\ \Leftrightarrow5x+4x-x=-5-15+12\\ \Leftrightarrow8x=-8\\ \Leftrightarrow x=-1\left(TM\right)\\ Vậy:S=\left\{-1\right\}\)

1 + 1 = 2

2 + 2 = 4

3 + 3 = 6 

:)))

1+1 = 2

2+2 = 4

3+3= 6

1+1=2

2+2=4

3+3=6

ok

         \(x^4+6x^3+11x^2+6x+1\)

\(=\left(x^4+6x^3+9x^2\right)+2\left(x^2+3x\right)+1\)

\(=\left(x^2+3x\right)^2+2\left(x^2+3x\right)+1\)

\(=\left(x^2+3x+1\right)^2\)

Chúc bạn học tốt.

a,17/6 b,12/7c,hình như bạn ghi sai đề

a)\(\frac{17}{6}\)

b)\(\frac{12}{7}\)

c)\(\frac{-11}{9}\)

Nhớ k cho mik nhé

a) 17/6

b) 24/14=12/7

c)-1/9

1/3 - 1/4 = 4/12 - 3/12 = 1/12

2/7 . 14/5 - 1 = 2.14/7.5 - 1 =4/5 - 1 = -1/5

3/4 - 1/25 x 5 = 3/4 - 1.5/25.1 = 3/4 - 1/5 = 11/20

-8 . ( 6. 1/24 ) = -8 . ( 6.1/24.1 ) = -8 . 1/4 = -8 . 1/4.1 = -2

1/3-1/4=1/12

2/7.14/5-1=-1/5

3/4-1/25.5=11/20

-8.(6.1/24)=-2 nhé

CHÚC BẠN HỌC GIỎI

K MÌNH NHÉ

a)1/12

b)-1/5

c)11/20

d)-2