Ôn tập toán 9 - Hỏi đáp

Nhận xét x > 0

Ta có : \(x^2=5+\sqrt{5+\sqrt{13+\sqrt{5+\sqrt{13+...}}}}\)

\(\Leftrightarrow x^2-5=\sqrt{13+\sqrt{5+\sqrt{13+....}}}\)

\(\Leftrightarrow\left(x^2-5\right)^2=13+\sqrt{5+\sqrt{13+...}}\)

\(\Leftrightarrow\left(x^2-5\right)^2-13=x\)

\(\Leftrightarrow x^4-10x^2-x+12=0\)

\(\Leftrightarrow\left(x-3\right)\left(x^3+3x^2-x-4\right)=0\)

Vì pt \(x^3+3x^2-x-4=0\) luôn có nghiệm \(x< 2\) mà \(x>\sqrt{5}>\sqrt{4}=2\)

Vậy x = 3

a) \(\dfrac{1}{\sqrt{3}}+\dfrac{1}{3\sqrt{2}}+\dfrac{1}{\sqrt{3}}\sqrt{\dfrac{5}{12}-\dfrac{1}{\sqrt{6}}}\)

\(=\dfrac{\sqrt{3}}{3}+\dfrac{\sqrt{2}}{6}+\dfrac{1}{\sqrt{3}}\sqrt{\dfrac{5}{12}-\dfrac{\sqrt{6}}{6}}\)

\(=\dfrac{\sqrt{3}}{3}+\dfrac{\sqrt{2}}{6}+\dfrac{\sqrt{\dfrac{5}{12}-\dfrac{\sqrt{6}}{6}}}{\sqrt{3}}\)

\(=\dfrac{\sqrt{3}}{3}+\dfrac{\sqrt{2}}{6}+\dfrac{\sqrt{\left(\dfrac{5}{12}-\dfrac{\sqrt{6}}{6}\right)\cdot3}}{3}\)

\(=\dfrac{\sqrt{3}}{3}+\dfrac{\sqrt{2}}{6}+\dfrac{\sqrt{\dfrac{5}{4}-\dfrac{\sqrt{6}}{2}}}{3}\)

\(=\dfrac{\sqrt{3}+\sqrt{\dfrac{5}{4}-\dfrac{\sqrt{6}}{2}}}{3}+\dfrac{\sqrt{2}}{6}\)

b) \(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}=...\)

c) \(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}=...\)

d) \(\dfrac{2\sqrt{3-\sqrt{3+\sqrt{13+\sqrt{48}}}}}{\sqrt{6}-\sqrt{2}}\)

\(=\dfrac{2\sqrt{3-\sqrt{3+\sqrt{13+4\sqrt{3}}}}}{\sqrt{6}-\sqrt{2}}\)

\(=\dfrac{2\sqrt{3-\sqrt{3+\sqrt{\left(1+2\sqrt{3}\right)^2}}}}{\sqrt{6}-\sqrt{2}}\)

\(=\dfrac{2\sqrt{3-\sqrt{3+1+2\sqrt{3}}}}{\sqrt{6}-\sqrt{2}}\)

\(=\dfrac{2\sqrt{3-\sqrt{3+2\sqrt{3}+1}}}{\sqrt{6}-\sqrt{2}}\)

\(=\dfrac{2\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}{\sqrt{6}-\sqrt{2}}\)

\(=\dfrac{2\sqrt{3-\left(\sqrt{3}+1\right)}}{\sqrt{6}-\sqrt{2}}\)

\(=\dfrac{2\sqrt{3-\left(\sqrt{3}+1\right)}\cdot\left(\sqrt{6}+\sqrt{2}\right)}{4}\)

\(=\dfrac{\sqrt{3\left(\sqrt{3}+1\right)}\cdot\left(\sqrt{6}+\sqrt{2}\right)}{2}\)

\(=\dfrac{\sqrt{3-\sqrt{3}-1}\sqrt{\left(\sqrt{6}+\sqrt{2}\right)^2}}{2}\)

\(=\dfrac{\sqrt{\left(3-\sqrt{3}-1\right)\cdot\left(\sqrt{6}+\sqrt{2}\right)^2}}{2}\)

\(=\dfrac{\sqrt{\left(2-\sqrt{3}\right)\cdot\left(6+2\sqrt{12}+2\right)}}{2}\)

\(=\dfrac{\sqrt{\left(2-\sqrt{3}\right)\cdot\left(6+4\sqrt{3}+2\right)}}{2}\)

\(=\dfrac{\sqrt{\left(2-\sqrt{3}\right)\cdot\left(8+4\sqrt{3}\right)}}{2}\)

\(=\dfrac{\sqrt{\left(2-\sqrt{3}\right)\cdot4\left(2+\sqrt{3}\right)}}{2}\)

\(=\dfrac{\sqrt{\left(4-3\right)\cdot4}}{2}\)

\(=\dfrac{\sqrt{1\cdot4}}{2}\)

\(=\dfrac{2}{2}\)

\(=1\)

1. \(\sqrt{7+2\sqrt{10}}-\sqrt{7-2\sqrt{10}}=\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}\\ =\sqrt{5}+\sqrt{2}-\sqrt{5}+\sqrt{2}=2\sqrt{2}\)

2. \(\sqrt{12-6\sqrt{3}}+\sqrt{21-12\sqrt{3}}=\sqrt{\left(3-\sqrt{3}\right)^2}+\sqrt{\left(2\sqrt{3}-3\right)^2}\\ =3-\sqrt{3}+2\sqrt{3}-3=\sqrt{3}\)

3. \(\sqrt{33-12\sqrt{6}}+\sqrt{15-6\sqrt{6}}=\sqrt{\left(2\sqrt{6}-3\right)^2}+\sqrt{\left(3+\sqrt{6}\right)^2}\\ =2\sqrt{6}-3+3+\sqrt{6}=3\sqrt{6}\)

1.\(\sqrt{7+2\sqrt{10}}-\sqrt{7-2\sqrt{10}}=\sqrt{\left(\sqrt{2}+\sqrt{5}\right)^2}-\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}\)

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

2. \(\sqrt{12-6\sqrt{3}+\sqrt{21-12\sqrt{3}}}=\sqrt{12-6\sqrt{3}+\sqrt{\left(3-2\sqrt{3}\right)^2}}\)

\(=\sqrt{12-6\sqrt{3}+2\sqrt{3}-3}=\sqrt{9-4\sqrt{3}}\)

3. \(\sqrt{33-12\sqrt{6}}+\sqrt{15-6\sqrt{6}}=\sqrt{\left(2\sqrt{6}-3\right)^2}+\sqrt{\left(\sqrt{6}-3\right)^2}\)

\(=2\sqrt{6}-3+3-\sqrt{6}=\sqrt{6}\)

a đưa về bình phương r` dùng cauchy, b cx v

Từ A kẻ đường cao AH (H thuộc BC) , Từ B kẻ đường cao BK (K thuộc AC)

Ta có : \(sinA=\frac{BK}{AB}\) ; \(sinB=\frac{AH}{AB}\) ; \(sinC=\frac{AH}{AC}\)

\(\Rightarrow\frac{AB}{sinC}=\frac{AB}{\frac{AH}{AC}}=\frac{AB.AC}{AH}\) ; \(\frac{AC}{sinB}=\frac{AC}{\frac{AH}{AB}}=\frac{AB.AC}{AH}\)

\(\Rightarrow\frac{c}{sinC}=\frac{b}{sinB}\) (1)

Lại có : \(BK=sinC.BC\Rightarrow\frac{BC}{sinA}=\frac{BC}{\frac{BK}{AB}}=\frac{BC.AB}{BK}=\frac{AB.BC}{sinC.BC}=\frac{AB}{sinC}\)

\(\Rightarrow\frac{a}{sinA}=\frac{c}{sinC}\) (2)

Từ (1) và (2) ta có : \(\frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}\) (Đpcm)

Do : Góc ABD = Góc ACE (= 90 - A )
=> Δ ABD  Δ ACE (2 Δ vuông)
=> AD.AC = AE.AB (tỉ lệ đồng dạng)
<=> AM^2 = AN^2 (Hệ thức lượng trong Δ vuông)
<=> AM = AN
Hay Δ AMN cân tại A.=>....

a) \(\sqrt{4-2\sqrt{3}}-\sqrt{3}=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}\)=\(\sqrt{3}-1-\sqrt{3}=-1\)

b) \(\sqrt{11+6\sqrt{2}}-3+\sqrt{2}\) = \(\sqrt{\left(3+\sqrt{2}\right)^2}-3+\sqrt{2}\)

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

c) \(\sqrt{9x^2}-2x=\sqrt{\left(3x\right)^2}-2x\) = \(\left|3x\right|-2x=-3x-2x\) (x < 0)

= \(-5x\)

d) \(x-4+\sqrt{16-8x+x^2}\) \(\left(x>4\right)\) = \(x-4+\sqrt{\left(4-x\right)^2}\)

= \(x-4+\left|4-x\right|\) = \(x-4-4+x\) ( \(x>4\))

= \(2x-8\)

a, \(\sqrt{4-2\sqrt{3}}-\sqrt{3}=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}=\sqrt{3}-1-\sqrt{3}=-1\)

b,\(\sqrt{11+6\sqrt{2}}-3+\sqrt{2}=\sqrt{\left(\sqrt{2}+3\right)^2}-3+\sqrt{2}=\sqrt{2}+3-3+\sqrt{2}=2\sqrt{2}\)

c, \(\sqrt{9x^2}-2x=\sqrt{\left(3x\right)^2}-2x=3x-2x=x\)

d, câu này sai đề rồi , nếu sửa lại phải như này :

\(x-4+\sqrt{16-8x+x^2}=x-4+\sqrt{\left(4-x\right)^2}=x-4+4-x=0\)

ta có : a) \(m>1\Leftrightarrow m^2>m\Leftrightarrow m^2>\left(\sqrt{m}\right)^2\Leftrightarrow m>\sqrt{m}\) (đpcm)

b) ta có \(m< 1\Leftrightarrow m^2< m\) (m là số dương ) \(\Leftrightarrow\) \(m^2< \left(\sqrt{m}\right)^2\)\(\Leftrightarrow\) \(m< \sqrt{m}\) (đpcm)

thay xyz=(4-x-y-z)²vào

1) \(P=\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)

\(=\left(4\sqrt{10}-4\sqrt{6}+\sqrt{150}-\sqrt{90}\right)\sqrt{4-\sqrt{15}}\)

\(=\left(4\sqrt{10}-4\sqrt{6}+5\sqrt{6}-3\sqrt{10}\right)\sqrt{4-\sqrt{15}}\)

\(=\left(\sqrt{10}+\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)

\(=\sqrt{\left(\sqrt{10}+\sqrt{6}\right)^2}\sqrt{4-\sqrt{15}}\)

\(=\sqrt{\left(\sqrt{10}+\sqrt{6}\right)^2+\left(4-\sqrt{15}\right)}\)

\(=\sqrt{\left(10+2\sqrt{60}+6\right)\cdot\left(4-\sqrt{15}\right)}\)

\(=\sqrt{\left(10+4\sqrt{15}+6\right)\cdot\left(4-\sqrt{15}\right)}\)

\(=\sqrt{\left(16+4\sqrt{15}\right)\cdot\left(4-\sqrt{15}\right)}\)

\(=\sqrt{4\left(4+\sqrt{15}\right)\cdot\left(4-\sqrt{15}\right)}\)

\(=\sqrt{4\left(16-15\right)}\)

\(=\sqrt{4\cdot1}\)

\(=\sqrt{4}\)

\(=2\)

2) \(Q=\left(3-\sqrt{5}\right)\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right)\sqrt{3-\sqrt{5}}\)

\(=\sqrt{\left(3-\sqrt{5}\right)^2}\sqrt{3+\sqrt{5}}+\sqrt{\left(3+\sqrt{5}\right)^2}\sqrt{3-\sqrt{5}}\)

\(=\sqrt{\left(3-\sqrt{5}\right)^2\cdot\left(3+\sqrt{5}\right)}+\sqrt{\left(3+\sqrt{5}\right)^2\cdot\left(3-\sqrt{5}\right)}\)

\(=\sqrt{\left(9-6\sqrt{5}+5\right)\cdot\left(3+\sqrt{5}\right)}+\sqrt{\left(9+6\sqrt{5}+5\right)\cdot\left(3-\sqrt{5}\right)}\)

\(=\sqrt{\left(14-6\sqrt{5}\right)\cdot\left(3+\sqrt{5}\right)}+\sqrt{\left(9+6\sqrt{5}+5\right)\cdot\left(3-\sqrt{5}\right)}\)

\(=\sqrt{42+14\sqrt{5}-18\sqrt{5}-30}+\sqrt{42-14\sqrt{5}+18\sqrt{5}-30}\)

\(=\sqrt{12-4\sqrt{5}}+\sqrt{12+4\sqrt{5}}\)

bài gì vậy trời ai mak siêng nổi

Bài quá dài

1) \(15-\sqrt{216}=15-\sqrt{4}.\sqrt{54}\)=\(9-2.\sqrt{9}.\sqrt{6}+6\)=\(\left(\sqrt{9}-\sqrt{6}\right)^2=\left(3-\sqrt{6}\right)^2\)

2)\(20-\sqrt{76}=20-\sqrt{4}.\sqrt{19}=19-2\sqrt{19}.1+1=\left(\sqrt{19}-1\right)^2\)

3)\(24-12\sqrt{3}=6\left(4-2\sqrt{3}\right)=6\left(3-2.\sqrt{3}.1+1\right)=6\left(\sqrt{3}-1\right)^2\)

4)\(7-\sqrt{13}=\frac{14-2\sqrt{13}}{2}=\frac{13-2\sqrt{13}.1+1}{2}=\frac{\left(\sqrt{13}-1\right)^2}{2}\)

5)\(16-\sqrt{31}=\frac{32-2\sqrt{31}}{2}=\frac{31-2\sqrt{31}.1+1}{2}=\frac{\left(\sqrt{31}-1\right)^2}{2}\)

a) \(\sqrt{3x^2-18x+28}+\sqrt{4x^2-24x+45}=-x^2+6x-5\) (ĐKXĐ : \(1\le x\le5\) )\

Ta có : \(\sqrt{3x^2-18x+28}+\sqrt{4x^2-24x+45}=\sqrt{3\left(x^2-6x+9\right)+1}+\sqrt{4\left(x^2-6x+9\right)+9}=\sqrt{3\left(x-3\right)^2+1}+\sqrt{4\left(x-3\right)^2+9}\)

\(\Rightarrow\sqrt{3x^2-18x+28}+\sqrt{4x^2-24x+45}\ge1+3=4\)

Lại có : \(-x^2+6x-5=-\left(x^2-6x+9\right)+4=-\left(x-3\right)^2+4\le4\)

Do đó, phương trình tương đương với : \(\begin{cases}1\le x\le5\\\sqrt{3x^2-18x+28}+\sqrt{4x^2-24x+45}=4\\-x^2+6x-5=4\end{cases}\)\(\Rightarrow x=3\left(TM\right)\)

Vậy nghiệm của phương trình là x = 3

b) \(\sqrt{x^2-4x+5}+\sqrt{x^2-4x+8}+\sqrt{x^2-4x+9}=3+\sqrt{5}\)

\(\Leftrightarrow\sqrt{\left(x-2\right)^2+1}+\sqrt{\left(x-2\right)^2+4}+\sqrt{\left(x-2\right)^2+5}=3+\sqrt{5}\)

Mặt khác, ta có : \(\begin{cases}\sqrt{\left(x-2\right)^2+1}\ge1\\\sqrt{\left(x-2\right)^2+4}\ge2\\\sqrt{\left(x-2\right)^2+5}\ge\sqrt{5}\end{cases}\)\(\Rightarrow\sqrt{x^2-4x+5}+\sqrt{x^2-4x+8}+\sqrt{x^2-4x+9}\ge3+\sqrt{5}\)

Dấu đẳng thức xảy ra <=> x = 2.

Vậy nghiệm của phương trình :  x = 2

b)\(\sqrt{x^2-4}-x^2+4=0\)

\(\Leftrightarrow\sqrt{x^2-4}=x^2-4\)

Đặt \(\sqrt{x^2-4}=t\) pt thành

\(t=t^2\Rightarrow t\left(1-t\right)=0\)

\(\Rightarrow\left[\begin{array}{nghiempt}t=1\\t=0\end{array}\right.\).

Với \(t=0\Rightarrow\sqrt{x^2-4}=0\Rightarrow x=\pm2\) 

Với \(t=1\Rightarrow\sqrt{x^2-4}=1\)\(\Rightarrow x=\pm\sqrt{5}\)

a)\(\sqrt{x^2-9}+\sqrt{x^2-6x+9}=0\)

\(\Rightarrow\sqrt{\left(x-3\right)\left(x+3\right)}+\sqrt{\left(x-3\right)^2}=0\)

\(\Rightarrow\sqrt{\left(x-3\right)\left(x+3\right)}+x-3=0\)

Đặt \(x-3=t\) pt thành

\(\sqrt{t\left(t-6\right)}-t=0\)

\(\Leftrightarrow t^2-6t=t^2\)

\(\Leftrightarrow t=0\)\(\Rightarrow x-3=0\Leftrightarrow x=3\)

a) \(\sqrt{\frac{2a^2b^4}{50}}=\sqrt{\frac{a^2b^4}{25}}=\frac{\sqrt{a^2b^4}}{\sqrt{25}}=\frac{ab^2}{5}\)

b) \(\frac{\sqrt{2ab^2}}{\sqrt{162}}=\sqrt{\frac{2ab^2}{162}}=\sqrt{\frac{ab^2}{81}}=\frac{\sqrt{ab^2}}{\sqrt{81}}=\frac{b\sqrt{a}}{9}\)