1) Đk: x \(\ge\)1

Ta có: \(\sqrt{3x-2}+\sqrt{x-1}=3\)

<=> \(3x-2+x-1+2\sqrt{\left(3x-2\right)\left(x-1\right)}=9\)

<=> \(2\sqrt{\left(3x-2\right)\left(x-1\right)}=12-4x\)

<=> \(\sqrt{3x^2-5x+2}=6-2x\)(\(1\le x\le3\))

<=> \(3x^2-5x+2=4x^2-24x+36\)

<=> \(x^2-19x+34=0\)

<=> \(x^2-17x-2x+34=0\)

<=> \(\left(x-17\right)\left(x-2\right)=0\)

<=> \(\orbr{\begin{cases}x=17\left(ktm\right)\\x=2\left(tm\right)\end{cases}}\)

Vậy S = {2}

3.Đk: x \(\ge\)0

 \(3x-7\sqrt{x}+4=0\)

<=> \(3x-3\sqrt{x}-4\sqrt{x}+4=0\)

<=> \(\left(3\sqrt{x}-4\right)\left(\sqrt{x}-1\right)=0\)

<=> \(\orbr{\begin{cases}x=\frac{16}{9}\\x=1\end{cases}}\left(tm\right)\)

4. Đk: x \(\ge\)0; x \(\ne\)16; x \(\ne\)49

Ta có: \(\frac{\sqrt{x}-2}{\sqrt{x}-4}=\frac{6-\sqrt{x}}{7-\sqrt{x}}\)

=> \(\left(\sqrt{x}-2\right)\left(\sqrt{x}-7\right)=\left(\sqrt{x}-4\right)\left(\sqrt{x}-6\right)\)

<=> \(x-9\sqrt{x}+14=x-10\sqrt{x}+24\)

<=> \(\sqrt{x}=10\) <=> x = 100 (tm)

5.Đk: x \(\ge\)1

 \(\frac{1}{2}\sqrt{x-1}-\frac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)

<=> \(-\sqrt{x-1}=-17\) <=> \(x-1=17\) <=> x = 18 (tm)

chỉ có max thôi bạn nhé

Ta có : \(x-5\sqrt{x}+2=x-2.\frac{5}{2}\sqrt{x}+\frac{25}{4}-\frac{17}{4}\)

\(=\left(\sqrt{x}+\frac{5}{2}\right)^2-\frac{17}{4}\ge-\frac{17}{4}\)

\(\frac{1}{\left(\sqrt{x}+\frac{5}{2}\right)^2-\frac{17}{4}}\le\frac{1}{-\frac{17}{4}}=-\frac{4}{17}\)

Dấu ''='' không xảy ra vì \(\sqrt{x}+\frac{5}{2}\ne0\)

ĐKXĐ : \(0\le x\le1\)

Bình phương 2 vế được : \(2x+2\sqrt{x^2-x^4}=x^2+2x+1\)  ( do x \(\ge0\Rightarrow x+1>0\) )

\(\Leftrightarrow2\sqrt{x^2-x^4}=x^2+1\Leftrightarrow4\left(x^2-x^4\right)=x^4+2x^2+1\) 

\(\Leftrightarrow5x^4-2x^2+1=0\) \(\Leftrightarrow PTVN\)

bn thử dùng bình phương xem nhưng nó quá dài và đặt ẩn phụ thì cũng vậy nên cách này là tối ưu nhất 

TXĐ : \(D=\left[-2;6\right]\)

\(VT=x^2-6x+13=\left(x-3\right)^2+4\ge4\forall x\)

\(VP=\sqrt{6-x}+\sqrt{x+2}\frac{\le}{B.C.S}\sqrt{\left(1+1\right)\left(6+2\right)}=4\)

\(\Rightarrow VT=VP=4\) 

" = " \(\Leftrightarrow x=3\) (t/m)

a) x = 25 => B = \(\frac{\sqrt{25}+3}{\sqrt{25}+1}=\frac{5+3}{5+1}=\frac{8}{6}=\frac{4}{3}\)

\(A=\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{1}{\sqrt{x}+2}-\frac{3\sqrt{x}}{x+\sqrt{x}-2}\)

\(A=\frac{\sqrt{x}\left(\sqrt{x}+2\right)+\sqrt{x}-1-3\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

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

3. \(S=A.B=\frac{\sqrt{x}+1}{\sqrt{x}+2}\cdot\frac{\sqrt{x}+3}{\sqrt{x}+1}=\frac{\sqrt{x}+3}{\sqrt{x}+2}=1+\frac{1}{\sqrt{x}+2}\)

Do \(\sqrt{x}+2\ge2\forall x\in R\) => \(\frac{1}{\sqrt{x}+2}\le\frac{1}{2}\) => \(S=1+\frac{1}{\sqrt{x}+2}\le1+\frac{1}{2}=\frac{3}{2}\)

Dấu "=" xảy ra<=> x=  0

Vậy MaxS = 3/2 <=> x=  0

1. x = 4 => B = \(\frac{3}{\sqrt{4}-1}=\frac{3}{2-1}=3\)

2. \(P=A-B=\frac{6}{x-1}+\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{3}{\sqrt{x}-1}\)

\(P=\frac{6+\sqrt{x}\left(\sqrt{x}-1\right)-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(P=\frac{6+x-\sqrt{x}-3\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(P=\frac{x-4\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{x-3\sqrt{x}-\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(P=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}-3}{\sqrt{x}+1}=\frac{\sqrt{x}+1-4}{\sqrt{x}+1}=1-\frac{4}{\sqrt{x}+1}\)

Do \(\sqrt{x}+1\ge1\forall x\in R\) => \(\frac{4}{\sqrt{x}+1}\le4\)=> \(1-\frac{4}{\sqrt{x}+1}\ge1-4=-3\)

=> \(P\ge-3\) => \(\frac{1}{P}\le-\frac{1}{3}\)

Dấu "=" xảy ra<=> x = 0

Vậy Max1/P = -1/3 <=> x = 6

a, Có ∠BAH= ∠BCA (vì cùng phụ với ∠HAC)

=> ∠BAH+ ∠HAD= ∠BCA + ∠DAC (vì AD là tia phân giác ∠HAC)

=> ∠BAD= ∠BCA + ∠DAC 

Xét ΔADC có ∠ADB là góc ngoài tại D => ∠ADB= ∠BCA + ∠DAC 

=> ∠BAD= ∠ADB

=> ΔABD cân tại B

b, Xét ΔABD cân tại B => AB= BD

Xét ΔABC vuông tại A

=> AB²= BH. BC

            = (BD- HD). BC

            = (AB- 6). 25

            = 25 AB- 150

=> AB²- 25AB+ 150= 0

<=> (AB-15)(AB-10)= 0 

<=> AB= 15 hoặc AB= 10

Vậy AB= 15cm, hoặc AB= 10 cm

Hình bạn tự vẽ nhé !

Xét tam giác ABC vuông tại A có đường cao AH

=> \(AB^2=BH.BC\) ( Hệ thức lượng trong tam giác vuông ) 

\(\Leftrightarrow BC=\frac{AB^2}{BH}=\frac{9^2}{5,4}=\frac{81}{5,4}=15\left(cm\right)\)

\(\Leftrightarrow CH=BC-BH=15-5,4=9,6\left(cm\right)\)

\(\Leftrightarrow AH^2=BH.CH\) ( Hệ thức lượng trong tam giác vuông )

\(\Leftrightarrow AH^2=5,4.9,6=51,84\Leftrightarrow AH=7,2\left(cm\right)\)

\(\Leftrightarrow AC^2=CH.BC\) ( Hệ thức lượng trong tam giác vuông ) 

\(\Leftrightarrow AC^2=15.9,6=144\Leftrightarrow AC=12\left(cm\right)\)

Đáp số : ...........

$\begin{array}{l} {x^3} + a{x^2} + bx + c = \left( {x + 1} \right)P\left( x \right) + 2021\\ \Rightarrow P\left( { - 1} \right) = 2021 \Rightarrow - 1 + a - b + c = 2021\\ {x^3} + a{x^2} + bx + c = \left( {x - 2} \right)P\left( x \right) + 2030\\ \Rightarrow P\left( 2 \right) = 2030 \Rightarrow 8 + 4a + 2b + c = 2030 \end{array}$

$\begin{array}{l} \Rightarrow \left\{ \begin{array}{l} 4a + 2b + c = 2022\\ a - b + c = 2022 \end{array} \right. \Rightarrow 4a + 2b + c = a - b + c\\ \Rightarrow 3a + 3b = 0 \Leftrightarrow a = - b\\ \Rightarrow K = \left( {{a^{2021}} + {b^{2021}}} \right)\left( {{a^{2022}} + {b^{2022}}} \right) = \left( {{a^{2021}} - {a^{2021}}} \right)\left( {{a^{2022}} + {b^{2022}}} \right)\\ = 0\left( {{a^{2022}} + {b^{2022}}} \right) = 0 \end{array}$

b) Đặt $n^2-n+5=k^2(k\in \mathbb Z)$

$\begin{array}{l} \Rightarrow 4{n^2} - 4n + 20 = 4{k^2}\\ \Rightarrow {\left( {2n - 1} \right)^2} + 19 = {\left( {2k} \right)^2}\\ \Rightarrow \left( {2k - 2n + 1} \right)\left( {2k + 2n - 1} \right) = 19 \end{array}$

$\begin{array}{l} k \in \mathbb Z,n \in \mathbb Z \to 2k - 2n + 1,2k + 2n - 1 \in \mathbb Z\\ \bullet \left\{ \begin{array}{l} 2k - 2n + 1 = 1\\ 2k + 2n - 1 = 19 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} 2k = 2n\\ 2n + 2n = 20 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} k = \dfrac{{20}}{3}\\ n = \dfrac{{10}}{3} \end{array} \right.\left( L \right) \end{array}$

$\begin{array}{l} \bullet \left\{ \begin{array}{l} 2k - 2n + 1 = - 1\\ 2k + 2n - 1 = - 19 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} 2k = 2n - 2\\ 2k + 2n = - 18 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} k = - 5\\ n = - 4 \end{array} \right.\left( {tm} \right)\\ \bullet \left\{ \begin{array}{l} 2k - 2n + 1 = 19\\ 2k + 2n - 1 = 1 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} k = 5\\ n = - 4 \end{array} \right.\left( {tm} \right)\\ \bullet \left\{ \begin{array}{l} 2k - 2n + 1 = - 19\\ 2k + 2n - 1 = - 1 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} k = - 5\\ n = 5 \end{array} \right.\left( {tm} \right) \end{array}$

Vậy $n=-4, n=5$ thỏa mãn yêu cầu bài toán.

Ta có : x² + y² \(\ge\)(x + y)²/2

=> (x + y)² \(\le\)2 => \(-\sqrt{2}\le x+y\le\sqrt{2}\)

Áp dụng bđt bunhiacopxki, ta có:

\(\left(x\sqrt{y+1}+y\sqrt{x+1}\right)^2\le\left(x^2+y^2\right)\left(y+1+x+1\right)\le\sqrt{2}+2\)

=> \(x\sqrt{y+1}+y\sqrt{y+1}\le\sqrt{\sqrt{2}+2}\)

Vậy Max \(x\sqrt{y+1}+y\sqrt{x+1}=\sqrt{2+\sqrt{2}}\) <=> \(x=y=\frac{\sqrt{2}}{2}\)