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`5)(6-sqrt6)/(1-sqrt6)+(6-sqrt6)/sqrt6=(sqrt6(sqrt6-1))/(1-sqrt6)+(sqrt6(sqrt6-1))/sqrt6=-sqrt6+sqrt6-1=-1` $\\$ `6)1/(sqrt2-sqrt3)-1/(sqrt3+sqrt2)=(sqrt2+sqrt3)/(2-3)-(sqrt3-sqrt2)/(3-2)=-(sqrt2+sqrt3)-sqrt3+sqrt2=-2sqrt3` $\\$ `7)1/(sqrt5+sqrt3)-1/(sqrt5-sqrt3)=(sqrt5-sqrt3)/(5-3)-(sqrt5+sqrt3)/(5-3)=(sqrt5-sqrt3-sqrt5-sqrt3)/2=-sqrt3` $\\$ `8)6/(1-sqrt3)-(3sqrt3-3)/(sqrt3+1)=(6(1+sqrt3))/(1-3)-(3(sqrt3-1)^2)/(3-1)=(-6(sqrt3+1)-3(4-2sqrt3))/2=-9`
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Bài 5:
b) Ta có: \(B=\dfrac{1+\sqrt{a}}{a\sqrt{a}+a+\sqrt{a}}:\dfrac{1}{a^2+\sqrt{a}}\)
\(=\dfrac{\sqrt{a}+1}{\sqrt{a}\left(a+\sqrt{a}+1\right)}\cdot\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{1}\)
\(=\dfrac{\left(\sqrt{a}+1\right)^2\cdot\left(a-\sqrt{a}+1\right)}{a+\sqrt{a}+1}\)
\(B=\dfrac{1+\sqrt{a}}{a\sqrt{a}+a+\sqrt{a}}:\dfrac{1}{a^2+\sqrt{a}}=\dfrac{1+\sqrt{a}}{\sqrt{a}\left(a+\sqrt{a}+1\right)}.\left[\sqrt{a}\left(a\sqrt{a}+1\right)\right]=\dfrac{\left(1+\sqrt{a}\right)\sqrt{a}.\left(\sqrt{a}+1\right)\left(\sqrt{a}-a+1\right)}{\sqrt{a}\left(a+\sqrt{a}+1\right)}=\dfrac{\left(1+\sqrt{a}\right)^2.\left(\sqrt{a}-a+1\right)}{a+\sqrt{a}+1}\)
P/s: Ko biết có sai đâu ko mà kết quả ra dài thek nhở ??
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\(\sqrt{15+5\sqrt{5}}-\sqrt{3-\sqrt{5}}=\sqrt{5}\sqrt{3+\sqrt{5}}-\sqrt{\dfrac{6-2\sqrt{5}}{2}}\)
\(=\sqrt{5}\sqrt{\dfrac{6+2\sqrt{5}}{2}}-\sqrt{\dfrac{\left(\sqrt{5}-1\right)^2}{2}}=\sqrt{5}\sqrt{\dfrac{\left(\sqrt{5}+1\right)^2}{2}}-\dfrac{\left|\sqrt{5}-1\right|}{\sqrt{2}}\)
\(=\sqrt{5}.\dfrac{\left|\sqrt{5}+1\right|}{\sqrt{2}}-\dfrac{\sqrt{5}-1}{\sqrt{2}}=\sqrt{5}.\dfrac{\sqrt{5}+1}{\sqrt{2}}-\dfrac{\sqrt{5}-1}{\sqrt{2}}\)
\(=\dfrac{5+\sqrt{5}-\sqrt{5}+1}{\sqrt{2}}=\dfrac{6}{\sqrt{2}}=3\sqrt{2}\)
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6:
1: BH=căn 15^2-12^2=9cm
BC=15^2/9=25cm
AC=căn 25^2-15^2=20cm
C ABC=15+20+25=60cm
XétΔHAB vuông tại H có sin BAH=BH/AB=9/15=3/5
nên góc BAH=37 độ
2: ΔABC vuông tại A có AH là đường cao
nên CA^2=CH*CB
ΔCAH vuông tại H có HF là đường cao
nên CF*CA=CA^2=CH*CB
3: Xét tứ giác AFHB có
HF//AB
góc AFH=90 độ
=>AFHB là hình thang vuông
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Áp dụng BĐT Bunhiacopxki cho 2 bộ số (\(\sqrt{a+b}\),\(\sqrt{b+c}\),\(\sqrt{a+c}\)) và (1,1,1) có: (1.\(\sqrt{a+b}\)+1.\(\sqrt{b+c}\)+1.\(\sqrt{a+c}\))2 ≤ (a + b + b + c + c + a)(12 + 12 + 12)
=> S2 ≤ 2.3 = 6 ⇔ S ≤ \(\sqrt{6}\)
Dấu "=" xảy ra ⇔ \(\sqrt{a+b}\) = \(\sqrt{b+c}\) = \(\sqrt{a+c}\) ⇔ a +b = b + c = c + a
⇔ 1 - c = 1 - a = 1 - b
⇔ a = b = c = \(\dfrac{1}{3}\)
Vậy maxS = \(\sqrt{6}\) ⇔ a = b = c = \(\dfrac{1}{3}\)
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Bài 5:
a. Khi $m=3$ thì hệ trở thành:
\(\left\{\begin{matrix} 3x+5y=7\\ 3x-y=1\end{matrix}\right.\Rightarrow (3x+5y)-(3x-y)=7-1=6\)
$\Leftrightarrow 6y=6$
$\Leftrightarrow y=1$
$x=\frac{y+1}{3}=\frac{1+1}{3}=\frac{2}{3}$
b.
Từ PT(2) suy ra $y=3x-1$. Thay vào PT(1) thì:
$mx+5(3x-1)=7$
$\Leftrightarrow x(m+15)=12(*)$
Để HPT ban đầu có nghiệm $(x,y)$ duy nhất thì pt $(*)$ phải có nghiệm $x$ duy nhất. Điều này xảy ra khi $m+15\neq 0\Leftrightarrow m\neq -15$
Khi đó:
$y=\frac{12}{m+15}$
$x=\frac{1}{3}(y+1)=\frac{1}{3}.\frac{m+27}{m+15}$
Khi đó:
$2x-3y=\frac{2(m+27)}{3(m+15)}-\frac{36}{m+15}=-2$
$\Leftrightarrow \frac{2m+54-108}{3(m+15)}=-2$
$\Leftrightarrow 2m-54=-6(m+15)$
$\Rightarrow m=-4,5$
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Bài 5 :
a, ĐKXĐ ; \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
Ta có : \(P=1:\left(\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(=1:\left(\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{x-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\)
\(=1:\left(\dfrac{x+2+x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\)
\(=1:\left(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\)
\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}\)
b, - Xét \(P-3=\dfrac{x+\sqrt{x}+1-3\sqrt{x}}{\sqrt{x}}=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}>0\)
\(\Rightarrow P>3\)
\(P=1:\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{x-1}\right)\) (Đk:\(x\ge0;x\ne1\))
\(=1:\left[\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}+1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]\)
\(=1:\left[\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right]\)
\(=1:\dfrac{x+2+x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=1:\dfrac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=1:\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}=\sqrt{x}+1+\dfrac{1}{\sqrt{x}}\)
b) Áp dụng AM-GM có:
\(\sqrt{x}+\dfrac{1}{\sqrt{x}}\ge2\sqrt{\sqrt{x}.\dfrac{1}{\sqrt{x}}}=2\)
Dấu "=" xảy ra khi \(\sqrt{x}=\dfrac{1}{\sqrt{x}}\Leftrightarrow x=1\left(ktm\right)\)
\(\Rightarrow\)Dấu "=" không xảy ra
\(\Rightarrow\sqrt{x}+\dfrac{1}{\sqrt{x}}>2\)\(\Rightarrow\sqrt{x}+1+\dfrac{1}{\sqrt{x}}>3\)
hay P>3
Vậy...
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Bài 4 Xét \(\Delta ABC\) vuông tại B
\(\widehat{BAC}+\widehat{ACB}=90^o\\ \Rightarrow\widehat{ACB}=90^o-30^o=60^o\)
Theo định lý sin
\(\dfrac{AB}{sinC}=\dfrac{BC}{sinA}\\ \Rightarrow BC=\dfrac{AB.sinA}{sinC}=\dfrac{2.sin30^o}{sin60^o}=\dfrac{2\sqrt{3}}{3}\)
Theo định lý Pytago :
\(AB^2+BC^2=AC^2\\ \Rightarrow AC=\sqrt{AB^2+BC^2}=\sqrt{2^2+\left(\dfrac{2\sqrt{3}}{3}\right)^2}=\dfrac{4\sqrt{3}}{3}\)
Bài 5
Chiều dài mặt phẳng nghiêng là :
\(5:sin36^o=8,5\left(m\right)\)
4:
góc BCA=90-30=60 độ
cos BAC=BA/CA
=>2/CA=cos30=căn 3/2
=>CA=4/căn 3(cm)
=>CB=1/2*4/căn 3=2/căn 3(cm)
Bài 5;
Gọi mp nghiêng là AB, chiều cao là AC
=>ΔACB vuông tại C có AC=5m và góc B=36 độ
ΔABC vuông tại C nên sin ABC=AC/AB
=>5/AB=sin36
=>AB=8,51(m)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 6:
a: \(Q=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{3}\)
\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
ĐKXĐ: \(x\ge2\)
\(B=\sqrt{x-2-2\sqrt{x-2}+1}+\sqrt{x-2-6\sqrt{x-2}+9}\)
\(=\sqrt{\left(\sqrt{x-2}-1\right)^2}+\sqrt{\left(3-\sqrt{x-2}\right)^2}\)
\(=\left|\sqrt{x-2}-1\right|+\left|3-\sqrt{x-2}\right|\)
\(\ge\left|\sqrt{x-2}-1+3-\sqrt{x-2}\right|=2\)
Vậy \(B_{min}=2\), dấu "=" xảy ra khi \(1\le\sqrt{x-2}\le3\Rightarrow3\le x\le11\)
Em cảm ơn ạ