Bài 11: Số vô tỉ. Khái niệm về căn bậc hai

Ta thấy: \(\left|x-y+3\right|\ge0\forall x,y\)

\(\Rightarrow\left|x-y+3\right|+3\ge3\forall x,y\)

\(\Rightarrow\sqrt{\left|x-y+3\right|+3}\ge\sqrt{3}\forall x,y\)

Lại có: \(\sqrt{y-1}\ge0\forall x;y\)

\(\Rightarrow VT=\sqrt{y-1}+\sqrt{\left|x-y+3\right|+3}\ge\sqrt{3}=VP\)

Đẳng thức xảy ra khi \(\left\{{}\begin{matrix}\sqrt{y-1}=0\\\sqrt{\left|x-y+3\right|+3}=\sqrt{3}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}y-1=0\\\left|x-y+3\right|+3=3\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}y=1\\x-y+3=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=-2\\y=1\end{matrix}\right.\)

Đặt A= 1/1.2 + 1/2.3 + 1/3.4+...+ 1/999.1000

=1-1/2+1/2-1/3+1/3-1/4+...+1/999-1/1000

=1-1/1000

=999/1000

\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+.....+\dfrac{1}{999.1000}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+.....+\dfrac{1}{999}-\dfrac{1}{1000}\)

\(=1-\dfrac{1}{1000}\)

\(=\dfrac{999}{1000}\)

\(A=\dfrac{\sqrt{x}}{\sqrt{x}+2}\)

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

Để \(A\in Z\) \(\Rightarrow\dfrac{2}{\sqrt{x}+2}\in Z\) mà \(\sqrt{x}\ge0\)

\(\Rightarrow\sqrt{x}+2\ge2\)

\(\Rightarrow\sqrt{x}+2=2\)

\(\Rightarrow\sqrt{x}=0\)

Vậy x = 0

x = 4

a) Theo bài ra ta có : \(x+y+z=49\)

\(\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{4z}{5}\Rightarrow\dfrac{12x}{18}=\dfrac{12y}{16}=\dfrac{12z}{15}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta được :

\(\dfrac{12x}{18}=\dfrac{12y}{16}=\dfrac{12z}{15}\\ =\dfrac{12x+12y+12z}{18+16+15}\\ =\dfrac{12\left(x+y+z\right)}{49}\\ =\dfrac{12\cdot49}{49}\\ =12\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{12x}{18}=12\Rightarrow12x=216\Rightarrow x=18\\\dfrac{12y}{16}=12\Rightarrow12y=192\Rightarrow y=16\\\dfrac{12z}{15}=12\Rightarrow12z=180\Rightarrow z=15\end{matrix}\right.\)

\(\text{Vậy }x=18\\ y=16\\ z=15\)

b) Theo bài ra ta có : \(2x+3y-z=50\)

\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\\ \Rightarrow\dfrac{2\left(x-1\right)}{4}=\dfrac{3\left(y-2\right)}{9}=\dfrac{z-3}{4}\\ \Rightarrow\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta được :

\(\dfrac{2x-2}{4}=\dfrac{3y-2}{9}=\dfrac{z-3}{4}=\\ \dfrac{\left(2x-2\right)+\left(3y-6\right)-\left(z-3\right)}{4+9-4}\\ =\dfrac{2x-2+3y-6-z+3}{9}\\ =\dfrac{\left(2x+3y-z\right)-\left(2+6-3\right)}{9}\\ =\dfrac{50-5}{9}\\ =\dfrac{45}{9}\\ =5\\ \)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{2x-2}{4}=5\Rightarrow2x-2=20\Rightarrow2x=22\Rightarrow x=11\\\dfrac{3y-6}{9}=5\Rightarrow3y-6=45\Rightarrow3y=51\Rightarrow y=17\\\dfrac{z-3}{4}=5\Rightarrow z-3=20\Rightarrow z=23\end{matrix}\right.\)

\(\text{Vậy }x=11\\ y=17\\ z=23\)

a) Ta có \(\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{4z}{5}\Rightarrow\dfrac{2x}{3.12}=\dfrac{3y}{4.12}=\dfrac{4z}{5.12}\)

\(\Rightarrow\)\(\dfrac{x}{18}=\dfrac{y}{16}=\dfrac{z}{15}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

\(\dfrac{x}{18}=\dfrac{y}{16}=\dfrac{z}{15}=\dfrac{x+y+z}{18+16+15}=\dfrac{49}{49}=1\)

Suy ra:

\(\dfrac{x}{18}=1\Rightarrow x=18.1=18\)

\(\dfrac{y}{16}=1\Rightarrow y=16.1=16\)

\(\dfrac{z}{15}=1\Rightarrow z=15.1=15\)

Vậy x = 18 ; y = 16 ; z = 15

a)Ta có :

4=\(\sqrt{16}\)>\(\sqrt{14}\)và \(\sqrt{33}\)>\(\sqrt{29}\)

Do đó 4 + √33 > √29 + √14

b)Ta có:

\(\sqrt{48}< \sqrt{49}=7\)

\(\sqrt{120}< \sqrt{121}=11\)

Do đó √48+ √120 < 11+7=18

c) Ta có:\(\sqrt{23}< \sqrt{25}=5\)

\(\sqrt{15}< \sqrt{16}=4\)

Do đó √23 +√15 < 5+4=9

Mặt khác \(\sqrt{91}< \sqrt{81}=9\)

Vậy √23 +√15 < \(\sqrt{91}\)

\(B=\sqrt{x+2017}+2018\)

a) Đề biểu thức trên có nghĩa thì:

\(x+2017\ge0\Rightarrow x\ge-2017\)

b) Với mọi \(x\ge-2017\) ta có:

\(\sqrt{x+2017}\ge0\)

\(\Rightarrow\sqrt{x+2017}+2018\ge2018\)

Dấu "=" xảy ra khi:

\(\sqrt{x+2017}=0\Rightarrow x=-2017\)

\(\Rightarrow MIN_B=2018\) khi \(x=-2017\)

b) rỗng

c) rỗng

d) N

e) N

f) Z

(cái nào nhỏ hơn thì lấy thôi)

1)

a) \(\sqrt{x+2}=\dfrac{5}{7}\)

-> x+2 = \(\left(\dfrac{5}{7}\right)^{^2}\)=\(\dfrac{25}{49}\)

-> x = \(\dfrac{25}{49}-2=-\dfrac{73}{49}\)

b) \(\sqrt{x+2}-8=1\)

-> \(\sqrt{x+2}=1+8=9\)

-> \(x+2=9^2=81\)

-> x = 81 -2 = 79

c) 4 - \(\sqrt{x-0,2}=0,5\)

-> \(\sqrt{x-0,2}=4-0,5=3,5\)

-> x - 0,2 = (3,5)² = 12,25

-> x = 12,25 +0,2 = 12,45

2) a)

Với mọi x thì: \(\sqrt{x+24}\ge0\)

=> \(\sqrt{x+24}+\dfrac{4}{7}\ge\dfrac{4}{7}\)

Dấu "=" xảy ra khi : x + 24 = 0 <=> x = -24

Vậy Min_A = \(\dfrac{4}{7}\) khi x = -24

Đề dọa trẻ con à

\(\sqrt{x^2\sqrt{x^4\sqrt{x^8\sqrt{x^{16}}}}}=\sqrt{3^{14}}\)

\(\Leftrightarrow\sqrt{x^2\sqrt{x^4\sqrt{x^8\cdot x^8}}}=\sqrt{3^{14}}\)

\(\Leftrightarrow\sqrt{x^2\sqrt{x^4\cdot x^8}}=\sqrt{3^{14}}\)

\(\Leftrightarrow\sqrt{x^2\cdot x^6}=\sqrt{3^{14}}\)

\(\Leftrightarrow x^4=2187\)\(\Rightarrow x=\pm3\sqrt[4]{27}\)

a) \(2\sqrt{a^2}=2\left|a\right|=2a\) (vì \(a\ge0\))

b) \(\sqrt{3a^2}=\left|a\right|\sqrt{3}=-a\sqrt{3}\) (vì \(a< 0\))

c) \(5\sqrt{a^4}=5\sqrt{\left(a^2\right)^2}=5\left|a^2\right|=5a^2\)

d) \(\dfrac{1}{3}\sqrt{c^6}=\dfrac{1}{3}\sqrt{\left(c^3\right)^2}=\dfrac{1}{3}\left|c^3\right|=\dfrac{1}{3}\left(-c^3\right)=-\dfrac{1}{3}c^3\) (vì \(c< 0\Rightarrow c^3< 0\))

\(a)2\sqrt{a^2}=2.\left|a\right|=2a\) ( vì \(a\ge0\) )

\(b)\sqrt{3a^2}=\left|a\right|\sqrt{3}=-a\sqrt{3}\) ( vì \(a< 0\) )

\(c)5\sqrt{a^4}=5\sqrt{\left(a^2\right)^2}=5\left|a^2\right|=5a^2\)

\(d)\dfrac{1}{3}\sqrt{c^6}=\dfrac{1}{3}\sqrt{\left(c^3\right)^2}=\dfrac{1}{3}\left|c^3\right|=\dfrac{1}{3}\left(-c^3\right)\) ( vì \(c< 0\Rightarrow c^3< 0\) )

Chúc bn học tốt!