Toán

12) \(\sqrt{11+2\sqrt{30}}=\sqrt{\left(\sqrt{6}\right)^2+2.\sqrt{6}.\sqrt{5}+\left(\sqrt{5}\right)^2}\)

\(=\sqrt{\left(\sqrt{6}+\sqrt{5}\right)^2}=\sqrt{6}+\sqrt{5}\)

\(=\sqrt{6+2\cdot\sqrt{6}\cdot\sqrt{5}+5}\)

\(=\sqrt{\left(\sqrt{6}+\sqrt{5}\right)^2}=\sqrt{6}+\sqrt{5}\)

AC=căn 5^2+12^2=13cm

=>C ABC=C DEF=5+12+13=18+12=30cm

=>DE=EF=DF=10cm

AC=căn 5^2+12^2=13cm

=>C ABC=C DEF=5+12+13=18+12=30cm

=>DE=EF=DF=10cm

\(sin47^o=\dfrac{63}{x}\\ \Rightarrow x=\dfrac{63}{sin47^o}\approx86,142\)

Hình đầu

Hình sau:

\(cos38^o=\dfrac{16}{x}\\ \Rightarrow x=\dfrac{16}{cos38^o}\approx20,304\)

9. A

D

Chọn C

D sai đó em

\(UCLN\left(42;30\right)=6\)

Số bi đỏ mỗi túi

\(42:6=7\) (viên bi)

Số bi vàng mỗi túi

\(30:6=5\) (viên bi)

ƯCLN(42;30)=6

=>Có thể chia được nhiều nhất là 6 túi

Khi đó, mỗi túi có 7 viên đỏ và 5 viên vàng

giúp tui mai tui nộp rùi

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

\(=a-c+d-c+a-d\)

\(=2a-2c\)

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

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

\(=-2b-2d\)

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

\(=ab-ac-ad-ab-ac+ad\)

\(=-2ac\)

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

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

\(=2a-2b+2c-2d\)

a) \(x-\sqrt{2x+3}=-2x\)

\(\Leftrightarrow\sqrt{2x+3}=x+2x\)

\(\Leftrightarrow\sqrt{2x+3}=3x\)

\(\Leftrightarrow2x+3=9x^2\)

\(\Leftrightarrow9x^2-2x-3=0\)

\(\Rightarrow\Delta=\left(-2\right)^2-4\cdot9\cdot\left(-3\right)=112>0\)

\(\Leftrightarrow\left[{}\begin{matrix}x_1=\dfrac{2+\sqrt{112}}{18}=\dfrac{1+2\sqrt{7}}{9}\\x_2=\dfrac{2-\sqrt{112}}{18}=\dfrac{1-2\sqrt{7}}{9}\end{matrix}\right.\)

b) \(\dfrac{1}{x}=1-\dfrac{1}{x+1}\) (ĐK: \(x\ne0,x\ne-1\))

\(\Leftrightarrow\dfrac{1}{x}+\dfrac{1}{x+1}=1\)

\(\Leftrightarrow\dfrac{x+1}{x\left(x+1\right)}+\dfrac{x}{x\left(x+1\right)}=1\)

\(\Leftrightarrow\dfrac{x+1+x}{x\left(x+1\right)}=1\)

\(\Leftrightarrow\dfrac{2x+1}{x^2+x}=1\)

\(\Leftrightarrow2x+1=x^2+1\)

\(\Leftrightarrow x^2-2x=0\)

\(\Leftrightarrow x\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x-2=0\end{matrix}\right.\)

\(\Leftrightarrow x=2\left(tm\right)\)

c) \(\dfrac{2}{\sqrt{x+3}}=\dfrac{1}{\sqrt{x^2-9}}\) (ĐK: \(x\ge3\))

\(\Leftrightarrow2\sqrt{x^2-2}=\sqrt{x+3}\)

\(\Leftrightarrow\sqrt{4\left(x^2-9\right)}=\sqrt{x+3}\)

\(\Leftrightarrow4\left(x^2-9\right)=x+3\)

\(\Leftrightarrow4x^2-36=x+3\)

\(\Leftrightarrow4x^2-x-36-3=0\)

\(\Leftrightarrow4x^2-x-39=0\)

\(\Rightarrow\Delta=\left(-1\right)^2-4\cdot4\cdot\left(-39\right)=625>0\)

\(\Leftrightarrow\left[{}\begin{matrix}x_1=\dfrac{1+\sqrt{625}}{8}=\dfrac{13}{4}\left(tm\right)\\x_2=\dfrac{1-\sqrt{625}}{8}=-3\left(ktm\right)\end{matrix}\right.\)

Ta có: \(M=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}=\dfrac{\sqrt{x}+2-3}{\sqrt{x}+2}=1-\dfrac{3}{\sqrt{x}+2}\)

Vì \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+2\ge2\Rightarrow\dfrac{3}{\sqrt{x}+2}\le\dfrac{3}{2}\)

\(\Rightarrow M=1-\dfrac{3}{\sqrt{x}+2}\ge1-\dfrac{3}{2}=-\dfrac{1}{2}\)

Dấu bằng xảy ra khi và chỉ khi: \(\sqrt{x}=0\Leftrightarrow x=0\)

KL:...

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