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\(a,ĐK:x>0;x\ne1\\ A=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-1}{\sqrt{x}}\\ b,A>0\Leftrightarrow\sqrt{x}-1>0\left(\sqrt{x}>1\right)\\ \Leftrightarrow x>1\)
a: \(A=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
Câu 26:
\(A=\left(2\sqrt{7}-2\sqrt{3}-\sqrt{7}\right)\sqrt{7}+2\sqrt{21}\\ =\left(\sqrt{7}-2\sqrt{3}\right)\sqrt{7}+2\sqrt{21}=7-2\sqrt{21}+2\sqrt{21}=7\\ B=\sqrt{\dfrac{3}{75}}=\sqrt{\dfrac{1}{25}}=\dfrac{1}{5}\)
Câu 27:
\(a,\Leftrightarrow\sqrt{x}=4\Leftrightarrow x=16\\ b,\Leftrightarrow\sqrt{\left(x-2\right)^2}=3\Leftrightarrow\left|x-2\right|=3\\ \Leftrightarrow\left[{}\begin{matrix}x=3+2=5\\x=-3+2=-1\end{matrix}\right.\)
Câu 28:
\(a,\left\{{}\begin{matrix}x=\dfrac{6^2}{3}=12\\y=\sqrt{x\left(x+3\right)}=\sqrt{12\cdot15}=10\sqrt{3}\end{matrix}\right.\\ b,\left\{{}\begin{matrix}BC=\dfrac{AC}{\sin B}=\dfrac{5}{\sin50}\approx6,5\left(cm\right)\\AB=\sqrt{BC^2-AC^2}\approx4,2\left(cm\right)\end{matrix}\right.\)
\(\text{Δ}=\left(-3\right)^2-4\cdot\left(2m+1\right)\)
=9-8m-4=-8m+5
Để phương trình có nghiệm kép thì -8m+5=0
hay m=5/8
Pt trở thành \(x^2-3x+\dfrac{9}{4}=0\)
hay x=3/2
`x^2 -x=12`
`<=>x^2 -x-12=0`
`<=> x^2+3x-4x-12=0`
`<=> x(x+3)-4(x+3)=0`
`<=>(x+3)(x-4)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=4\end{matrix}\right.\)
`---`
`2x^2-3x=15-4x`
`<=> 2x^2-3x+4x=15`
`<=>2x^2 +x-15=0`
`<=>2x^2+6x-5x-15=0`
`<=> 2x(x+3)-5(x+3)=0`
`<=>(x+3)(2x-5)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{5}{2}\end{matrix}\right.\)
`---`
`x(x-5)=24`
`<=> x^2 -5x-24=0`
`<=>x^2+3x-8x-24=0`
`<=>x(x+3) -8(x+3)=0`
`<=>(x+3)(x-8)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=8\end{matrix}\right.\)
`----`
`x(x-3)=10(x-4)`
`<=> x^2 -3x =10x -40`
`<=>x^2 -3x-10x +40=0`
`<=> x^2 -13x+40=0`
`<=>x^2-5x-8x+40=0`
`<=> x (x-5) - 8(x-5)=0`
`<=>(x-5)(x-8)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=8\end{matrix}\right.\)
5. \(x^2-x=12\Leftrightarrow x^2-x-12=0\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)
6. \(2x^2-3x=15-4x\Leftrightarrow2x^2+x-15=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-3\end{matrix}\right.\)
7. \(x\left(x-5\right)=24\Leftrightarrow x^2-5x-24=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-3\end{matrix}\right.\)
8. \(x\left(x-3\right)=10\left(x-4\right)\Leftrightarrow x^2-3x=10x-40\)
\(\Leftrightarrow x^2-13x+40=0\Leftrightarrow\left[{}\begin{matrix}x=8\\x=5\end{matrix}\right.\)
Xét tam giác ABC vuông tại A có:
AC = AB.tan\(\widehat{ABC}\) = 4,5467. tan59o12'37'' \(\approx\) 7,6303 (cm)
BC = \(\dfrac{AB}{cos\widehat{ABC}}=\dfrac{4,5467}{cos59^o12'37"}\)\(\approx\) 8,8822 (cm)
Do AM là trung tuyến tam giác ABC
=> AM = MC = MB = \(\dfrac{1}{2}\)AB \(\approx\) \(\dfrac{1}{2}\).8,8822 = 4,4411 (cm)
Kẻ NK ⊥ AC => NK // AB (cùng vuông góc AC)
AN là phân giác trong tam giác ABC =>\(\widehat{NAB}=\widehat{NAC}=45^o\); \(\dfrac{NC}{NB}=\dfrac{AC}{AB}=\dfrac{7,6303}{4,5467}\) => \(\dfrac{NC}{BC}=\dfrac{7,6303}{12,177}\) (*)
Do NK // AB (cmt) => \(\dfrac{NK}{AB}=\) \(\dfrac{NC}{BC}=\dfrac{7,6303}{12,177}\)
=> NK = \(\dfrac{7,6303}{12,177}.AB=\dfrac{7,6303}{12,177}.4,5467\approx2,849\) (cm)
Xét tam giác ANK vuông tại K có: AN = \(\dfrac{NK}{sin\widehat{NAK}}=\dfrac{2,849}{sin45^o}\approx2,015\left(cm\right)\)
Kẻ AH ⊥ BC. Xét tam giác ABC vuông tại A có: AH.BC = AB.AC
=> AH = \(\dfrac{AB.AC}{BC}=\dfrac{4,5467.7,6303}{8,8822}\approx3,9509\left(cm\right)\)
Từ (*) => NC = \(\dfrac{7,6303}{12,177}.BC=\dfrac{7,6303}{12,177}.8,8822\approx5,5657\) (cm)
=> MN = NC - MC = 5,5657 - 4,4411 = 1,1246 (cm)
=> SAMN = \(\dfrac{1}{2}\).AH.MN = \(\dfrac{1}{2}\).3,9509.1,1246 \(\approx\) 2,2216 (cm2)
b: \(B=\dfrac{x-2\sqrt{x}+1-x-2\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{1}{\sqrt{x}}=\dfrac{-4}{x-1}\)