95%

\(\sqrt{\dfrac{2}{5}}\) + \(\sqrt{\dfrac{5}{2}}\) = \(\dfrac{\sqrt{2}\times\sqrt{2}}{\sqrt{10}}\)+ \(\dfrac{\sqrt{5}\times\sqrt{5}}{\sqrt{10}}\) = \(\dfrac{7}{\sqrt{10}}\)= \(\dfrac{7\sqrt{10}}{10}\)

Lời giải:

\(\sqrt{\frac{2}{5}}+\sqrt{\frac{5}{2}}=\frac{\sqrt{2}.\sqrt{2}+\sqrt{5}.\sqrt{5}}{\sqrt{5}.\sqrt{2}}=\frac{7}{\sqrt{10}}=\frac{7\sqrt{10}}{10}\)

a) \((2+\sqrt{3})\sqrt{7-4\sqrt{3}}\)

\(=(2+\sqrt{3})\sqrt{4-4\sqrt{3}+3}\)

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

\(=(2+\sqrt{3})(2-\sqrt{3})=4-3=1\)

b) \(\sqrt{(1-\sqrt{2023})^2}\cdot\sqrt{2024+2\sqrt{2023}}\)

\(=|1-\sqrt{2023}|\sqrt{2023+2\sqrt{2023}+1}\)

\(=(\sqrt{2023}-1)\sqrt{(\sqrt{2023}+1)^2}\)

\(=(\sqrt{2023}-1)(\sqrt{2023}+1)\)

\(=\sqrt{2023^2}-1^2=2023-1=2022\)

`1/2(\sqrt{6}+\sqrt{5})^2-1/4\sqrt{120}-\sqrt{15}/2`

`=1/2 .(6+5+2\sqrt{30})-1/4 \sqrt{2^2 .30}-\sqrt{15}/2`

`=11/2+\sqrt{30}-1/2 \sqrt{30}-\sqrt{15}/2`

`=[11+2\sqrt{30}-\sqrt{30}-\sqrt{15}]/2`

`=[11+\sqrt{30}-\sqrt{15}]/2`

Lời giải:

$\frac{1}{2}(\sqrt{6}+\sqrt{5})^2-\frac{1}{4}\sqrt{120}-\frac{\sqrt{15}}{2}$

$=\frac{1}{2}(11+2\sqrt{30})-\frac{1}{2}\sqrt{30}-\frac{\sqrt{15}}{2}$

$=\frac{11}{2}+\frac{\sqrt{30}}{2}-\frac{\sqrt{15}}{2}$

a, △ABC vuông tại A có AH là đường cao.

\(\Rightarrow\left\{{}\begin{matrix}HB.BC=AB^2\\HC.BC=AC^2\end{matrix}\right.\) (hệ thức lượng trong tam giác vuông)

\(\Rightarrow\dfrac{HB.BC}{HC.BC}=\dfrac{HB}{HC}=\dfrac{AB^2}{AC^2}\)

b, △ABH vuông tại H có HD là đường cao.

\(\Rightarrow BD.AB=BH^2\) (hệ thức lượng trong tam giác vuông)

\(\Rightarrow BD=\dfrac{BH^2}{AB}\left(1\right)\)

△ACH vuông tại H có HE là đường cao.

\(\Rightarrow EC.AC=CH^2\) (hệ thức lượng trong tam giác vuông)

\(\Rightarrow EC=\dfrac{CH^2}{AC} \left(2\right)\)

Từ (1), (2) suy ra:

\(\dfrac{DB}{EC}=\dfrac{\dfrac{BH^2}{AB}}{\dfrac{CH^2}{AC}}=\left(\dfrac{BH}{CH}\right)^2.\dfrac{AC}{AB}=\dfrac{AB^4}{AC^4}.\dfrac{AC}{AB}=\left(\dfrac{AB}{AC}\right)^3\)

c, Có: \(\left\{{}\begin{matrix}BD.AB=BH^2\\EC.AC=CH^2\end{matrix}\right.\Rightarrow BD.EC.AB.AC=BH^2.CH^2\)

Mà \(\left\{{}\begin{matrix}BH.CH=AH^2\\AH.BC=AB.AC\end{matrix}\right.\)

\(\Rightarrow BD.EC.AH.BC=AH^4\)

\(\Rightarrow BD.EC.BC=AH^3\)

You yourself draw the figure.

a) Consider the right triangle ABC (which has \(\widehat{A}=90^o\)) has the height AH, thus, we have \(AB^2=HB.BC\) 

Similarly, we have \(AC^2=HC.BC\) 

From these, we get \(\dfrac{HB.BC}{HC.BC}=\dfrac{AB^2}{AC^2}\Leftrightarrow\dfrac{HB}{HC}=\left(\dfrac{AB}{AC}\right)^2\)

b) We can easily prove that \(\Delta BDH~\Delta HEC\left(a.a\right)\), therefore, \(\dfrac{DB}{HE}=\dfrac{HB}{HC}\)

Then, we can see that \(\dfrac{HB}{HC}=\left(\dfrac{AB}{AC}\right)^2\), so, we have \(\dfrac{DB}{HE}=\left(\dfrac{AB}{AC}\right)^2\), and the thing we have to prove is the same of \(\dfrac{DB}{HE}=\dfrac{DB}{EC}\) or \(HE=EC\), but this is clearly wrong. You have to edit the title.

c) This title is also wrong. \(BD.CE.BC=DB^3\Leftrightarrow CE.BC=DB^2\) which make no sense.

a, Với x khác 1 ; x khác 2/3 

\(C=\left(\dfrac{2x}{2x^2-5x+3}-\dfrac{5}{2x-3}\right):\left(3+\dfrac{2}{1-x}\right)\)

\(=\left(\dfrac{2x-5x+5}{\left(2x-3\right)\left(x-1\right)}\right):\left(\dfrac{3-3x+2}{1-x}\right)=\dfrac{\left(5-3x\right)\left(1-x\right)}{\left(2x-3\right)\left(x-1\right)\left(5-3x\right)}=\dfrac{1}{3-2x}\)

b, Ta có |2x-3| + 1 = 8 <=> | 2x - 3 | = 7 

TH1 : 2x - 3 = 7 <=> x = 5 

TH2 : 2x - 3 = -7 <=> x = -2 

Thay x = 5 vào ta được \(\dfrac{1}{3-2.5}=-\dfrac{1}{7}\)

Thay x = -2 vào ta được \(\dfrac{1}{3-2\left(-2\right)}=\dfrac{1}{7}\)

c, Ta có \(\dfrac{1}{3-2x}>0\Rightarrow3-2x>0\Leftrightarrow x< \dfrac{3}{2}\)

d, 3 - 2x thuộc Ư(1) = {-1;1}

e, \(\dfrac{1}{3-2x}=\dfrac{1}{6-x^2}\Leftrightarrow x^2-2x-3=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)

\(A=\left(\dfrac{\sqrt{x}-1}{x-1}+\dfrac{2-2\sqrt{x}}{x\sqrt{x}+x-\sqrt{x}-1}\right)\div\left(\dfrac{\sqrt{x}+2}{x+\sqrt{x}-2}-\dfrac{2}{x-1}\right)\)

\(=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2}{\left(\sqrt{x}+1\right)^2}\right)\div\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{x-1}\right)\)

\(=\dfrac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)^2}\div\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)

bạn thêm đk a > 0 nhé 

Xét tam giác ABC vuông tại B

 Theo định lí Pytago ta được 

\(AC=\sqrt{AB^2+BC^2}=\sqrt{a^2+3a^2}=\left|2a\right|=2a\)

Áp dụng hệ thức \(\dfrac{1}{BH^2}=\dfrac{1}{AB^2}+\dfrac{1}{BC^2}=\dfrac{1}{a^2}+\dfrac{1}{3a^2}=\dfrac{4a^2}{3a^4}\Rightarrow BH=\dfrac{\sqrt{3}a^2}{2a}=\dfrac{\sqrt{3}a}{2}\)

Áp dụng hệ thức \(AB^2=AH.AC\Rightarrow AH=\dfrac{AB^2}{AC}=\dfrac{a^2}{2a}=\dfrac{a}{2}\)

HC = AC - AH = \(2a-\dfrac{a}{2}=\dfrac{3a}{2}\)

Thế n = 2 ta được:

\(\dfrac{2.3.5}{6}=5\)

Xong

Xét \(\Delta AEH\) và \(\Delta AHB\), ta có:

\(\widehat{A}\) chung

\(\widehat{AEH}=\widehat{AHB}=90^o\)

\(\Rightarrow\Delta AEH~\Delta AHB\left(g.g\right)\)

\(\Rightarrow\dfrac{AE}{AH}=\dfrac{AH}{AB}\)

\(\Rightarrow AE.AB=AH^2\) (*)

Xét \(\Delta AFH\) và \(\Delta AHC\), ta có:

\(\widehat{A}\) chung

\(\widehat{AFH}=\widehat{AHC}=90^o\)

\(\Rightarrow\Delta AFH~\Delta AHC\left(g.g\right)\)

\(\Rightarrow\dfrac{AF}{AH}=\dfrac{AH}{AC}\)

\(\Rightarrow AF.AC=AH^2\) (**)

Từ (*)(**)\(\Rightarrow AE.AB=AF.AC\)