ai nay dung kinh nghiem la chinh

cau a)

ta thay \(10+6\sqrt{3}=\left(1+\sqrt{3}\right)^3\)

\(6+2\sqrt{5}=\left(1+\sqrt{5}\right)^2\)

khi do \(x=\frac{\sqrt[3]{\left(\sqrt{3}+1\right)^3}\left(\sqrt{3}-1\right)}{\sqrt{\left(1+\sqrt{5}\right)^2}-\sqrt{5}}\)

\(x=\frac{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}{1+\sqrt{5}-\sqrt{5}}\)

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

suy ra 

x^3-4x+1=1

A=1^2018

A=1

b)

ta thay

\(7+5\sqrt{2}=\left(1+\sqrt{2}\right)^3\)

khi do 

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

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

x=2

thay vao

x^3+3x-14=0

B=0^2018

B=0

28. \(x^2+\frac{9x^2}{\left(x-3\right)^2}=40\) DK: \(x\ne3\)

PT\(\Leftrightarrow\left(x+\frac{3x}{x-3}\right)^2-6\frac{x^2}{x-3}-40=0\)\(\Leftrightarrow\frac{x^4}{\left(x-3\right)^2}-6\frac{x^2}{x-3}-40=0\)

Dat \(\frac{x^2}{x-3}=a\). PTTT \(a^2-6a-40=0\)\(\Leftrightarrow\left(a-10\right)\left(a+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=10\\a=-4\end{matrix}\right.\)

giai tiep

14. \(\frac{1}{\sqrt{x}+1}+\frac{1}{\sqrt{x}-1}=1\) DK: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

PT\(\Leftrightarrow\frac{\sqrt{x}-1+\sqrt{x}+1}{x-1}=1\Leftrightarrow2\sqrt{x}=x-1\)\(\Leftrightarrow x-2\sqrt{x}+1=2\Leftrightarrow\left(\sqrt{x}-1\right)^2=2\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3+2\sqrt{2}\\x=3-2\sqrt{2}\end{matrix}\right.\)

Bạn ơi lần sau bạn đăng bài thì cố gắng đăng giãn giãn bớt bớt/ chia nhỏ bài ra chứ một cục bài như thế này nhìn rất đáng sợ và gây tâm lý ngại đọc nhé.

1. ĐKXĐ: $x\geq 1$

Đặt $x\sqrt{x-1}=a\Rightarrow x^3-x^2=x^2(x-1)=a^2$. PT đã cho trở thành:

$a^2+12a+20=0(*)$

Lại thấy rằng vì $x\geq 1$ nên $a\geq 0$

$\Rightarrow a^2+12a+20\geq 20>0$. Do đó $(*)$ vô nghiệm. Kéo theo PT ban đầu vô nghiệm.

2. ĐK: $x\geq -1$

$x^3+\sqrt{(x+1)^3}=9x+8$

$\Leftrightarrow x^3-9x-8+\sqrt{(x+1)^3}=0$

$\Leftrightarrow (x^2-x-8)(x+1)+(x+1)\sqrt{x+1}=0$

$\Leftrightarrow (x+1)(x^2-x-8+\sqrt{x+1})=0$

Nếu $x+1=0\Rightarrow x=-1$ (thỏa mãn)

Nếu $x^2-x-8+\sqrt{x+1}=0$

$\Leftrightarrow (x^2-9)-(x-3)+(\sqrt{x+1}-2)=0$

$\Leftrightarrow (x-3)\left(x+3+\frac{1}{\sqrt{x+1}+2}}-1\right)=0$

Dễ thấy với $x\geq -1$ thì biểu thức trong ngoặc lớn luôn lớn hơn $0$

Do đó $x-3=0\Rightarrow x=3$

Vậy $x=-1$ hoặc $x=3$

Bài 1:

a. Ta có \(\sqrt{\dfrac{2}{x^2}}=\dfrac{\sqrt{2}}{\left|x\right|}=\dfrac{\sqrt{2}}{x}\) ,để biểu thức có nghĩa thì \(x>0\)

b. Để biểu thức \(\sqrt{\dfrac{-3}{3x+5}}\) có nghĩa thì \(\dfrac{-3}{3x+5}\ge0\) 

mà \(-3< 0\Rightarrow3x+5< 0\) \(\Rightarrow x< \dfrac{-5}{3}\)

Bài 2:

a. \(\dfrac{2+\sqrt{2}}{1+\sqrt{2}}=\dfrac{\left(2+\sqrt{2}\right)\left(1-\sqrt{2}\right)}{1-2}=\dfrac{-\sqrt{2}}{-1}=\sqrt{2}\)

b. \(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right)\sqrt{7}+7\sqrt{8}\)

\(=14-14\sqrt{2}+7+14\sqrt{2}\)

\(=21\)

c. \(\left(\sqrt{14}-3\sqrt{2}\right)^2+6\sqrt{28}\)

\(=14-6\sqrt{28}+18+6\sqrt{28}\)

\(=32\)

1, \(x^3=\left(7+\sqrt{\frac{49}{8}}\right)+\left(7-\sqrt{\frac{49}{8}}\right)+3x\sqrt[3]{\left(7+\sqrt{\frac{49}{8}}\right)\left(7-\sqrt{\frac{49}{8}}\right)}\)

\(=14+3x\cdot\frac{7}{2}=14+\frac{21x}{2}\)

\(\Leftrightarrow x^3-\frac{21}{2}x-14=0\)

Ta có: \(f\left(x\right)=\left(2x^3-21-29\right)^{2019}=\left[2\left(x^3-\frac{21}{2}x-14\right)-1\right]^{2019}=\left(-1\right)^{2019}=-1\)

2, ta có: \(1^3+2^3+...+n^3=\left(1+2+...+n\right)^2=\left[\frac{n\left(n+1\right)}{2}\right]^2\) (bạn tự cm)

Áp dụng công thức trên ta được n=2016

3, \(x=\frac{\sqrt[3]{17\sqrt{5}-38}\left(\sqrt{5}+2\right)}{\sqrt{5}+\sqrt{14-6\sqrt{5}}}=\frac{\sqrt[3]{\left(\sqrt{5}\right)^3-3.\left(\sqrt{5}\right)^2.2+3\sqrt{5}.2^2-2^3}\left(\sqrt{5}+2\right)}{\sqrt{5}+\sqrt{9-2.3\sqrt{5}+5}}\)

\(=\frac{\sqrt[3]{\left(\sqrt{5}-2\right)^3}\left(\sqrt{5}+2\right)}{\sqrt{5}+\sqrt{\left(3-\sqrt{5}\right)^2}}=\frac{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}{\sqrt{5}+3-\sqrt{5}}=\frac{5-4}{3}=\frac{1}{3}\)

Thay x=1/3 vào A ta được;

\(A=3x^3+8x^2+2=3.\left(\frac{1}{3}\right)^3+8.\left(\frac{1}{3}\right)^2+2=3\)

Bài 4

ÁP DỤNG BĐT CAUCHY 

là ra

\(\frac{1}{\sqrt{1^3+2^3}}+\frac{1}{\sqrt{1^3+2^3+3^3}}+...+\frac{1}{\sqrt{1^3+2^3+3^3+...+n^3}}=\frac{2015}{2017}\) (1) 

Cần CM: \(1^3+2^3+3^3+...+n^3=\left(1+2+3+...+n\right)^2\) quy nạp nhé bn, trên mạng có nhìu 

(1) \(\Leftrightarrow\)\(\frac{1}{\sqrt{\left(1+2\right)^2}}+\frac{1}{\sqrt{\left(1+2+3\right)^2}}+...+\frac{1}{\sqrt{\left(1+2+3+...+n\right)^2}}=\frac{2015}{2017}\)

\(\Leftrightarrow\)\(\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+n}=\frac{2015}{2017}\)

\(\Leftrightarrow\)\(\frac{1}{\frac{2\left(2+1\right)}{2}}+\frac{1}{\frac{3\left(3+1\right)}{2}}+...+\frac{1}{\frac{n\left(n+1\right)}{2}}=\frac{2015}{2017}\)

\(\Leftrightarrow\)\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}\right)=\frac{2015}{2017}\)

\(\Leftrightarrow\)\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}\right)=\frac{2015}{2017}\)

\(\Leftrightarrow\)\(2\left(\frac{1}{2}-\frac{1}{n+1}\right)=\frac{2015}{2017}\)

\(\Leftrightarrow\)\(n=2016\)

a) \(\text{Đ}K\text{X}\text{Đ}:\frac{3}{2}\le x\le\frac{5}{2}\)

Áp dụng BĐT Bunhiacopxki ta có:

\(VT=\sqrt{2x-3}+\sqrt{5-2x}\le\sqrt{2\left(2x-3+5-2x\right)}=2\)

Dấu '=' xảy ra khi \(\sqrt{2x-3}=\sqrt{5-2x}\Leftrightarrow x=2\)

Lại có: \(VP=3x^2-12x+14=3\left(x-2\right)^2+2\ge2\)

Dấu '=' xảy ra khi x=2

Do đó VT=VP khi x=2

b) ĐK: \(x\ge0\). Ta thấy x=0 k pk là nghiệm của pt, chia 2 vế cho x ta có:

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

\(\Leftrightarrow\left(x+\frac{4}{x}\right)-\left(\sqrt{x}+\frac{2}{\sqrt{x}}\right)-2=0\)

Đặt \(\sqrt{x}+\frac{2}{\sqrt{x}}=t>0\Leftrightarrow t^2=x+4+\frac{4}{x}\Leftrightarrow x+\frac{4}{x}=t^2-4\), thay vào ta có:

\(\left(t^2-4\right)-t-2=0\Leftrightarrow t^2-t-6=0\Leftrightarrow\left(t-3\right)\left(t+2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}t=3\\t=-2\end{cases}}\)

Đối chiếu ĐK  của t

\(\Rightarrow t=3\Leftrightarrow\sqrt{x}+\frac{2}{\sqrt{x}}=3\Leftrightarrow x-3\sqrt{x}+2=0\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=1\end{cases}}\)

c) Đặt \(y=\sqrt{x^2+7x+7};y\ge0\)

Pt có dạng: \(3y^2+2y-5=0\Leftrightarrow\orbr{\begin{cases}y=\frac{-5}{3}\\y=1\end{cases}\Leftrightarrow y=1}\)

Với y=1\(\Leftrightarrow\sqrt{x^2+7x+7}=1\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-6\end{cases}}\)

Tiếp =))

c)Áp dụng BĐT AM-GM ta có:

\(x\sqrt{y-1}\le\frac{x\left(y-1+1\right)}{2}=\frac{xy}{2}\)

\(2y\sqrt{x-1}\le\frac{2y\left(x-1+1\right)}{2}=\frac{2xy}{2}\)

Cộng theo vế 2 BĐT trên ta có:

\(VT=x\sqrt{y-1}+2y\sqrt{x-1}\le\frac{3xy}{2}=VP\)

Nên xảy ra khi \(x=y\) thay vào giải ra có: x=y=2

d)\(\sqrt{2x^2+x+1}+\sqrt{x^2-x+1}=3x\)

\(pt\Leftrightarrow\sqrt{2x^2+x+1}-2+\sqrt{x^2-x+1}-1=3x-3\)

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

\(\Leftrightarrow\frac{\left(x-1\right)\left(2x+3\right)}{\sqrt{2x^2+x+1}+2}+\frac{x\left(x-1\right)}{\sqrt{x^2-x+1}+1}-3\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(\frac{\left(2x+3\right)}{\sqrt{2x^2+x+1}+2}+\frac{x}{\sqrt{x^2-x+1}+1}-3\right)=0\)

pt trong ngoặc vn nên x=1

Tắm đã làm nốt cho :))

Chả ai giúp t gank =)), mà lần sau đăng ít 1 thôi đăng lắm thế này nhìn nản cmn luôn ấy

a)\(\sqrt{x^2+x-5}+\sqrt{-x^2+x+3}=x^2-3x+4\)

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

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

\(\Leftrightarrow\frac{\left(x-2\right)\left(x+3\right)}{\sqrt{x^2+x-5}+1}+\frac{-\left(x-2\right)\left(x+1\right)}{\sqrt{-x^2+x+3}+1}-\left(x-1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left[\frac{\left(x+3\right)}{\sqrt{x^2+x-5}+1}-\frac{\left(x+1\right)}{\sqrt{-x^2+x+3}+1}-\left(x-1\right)\right]=0\)

Pt trong ngoặc <0 nên x=2 là nghiệm

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

Đk:\(x\ge-\frac{1}{2}\)

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

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

\(\Leftrightarrow\frac{x^2-3x}{2}-\frac{2x^3-5x^2-3x}{\sqrt{2x^3-x^2+x+1}+2x+1}=0\)

\(\Leftrightarrow\frac{x\left(x-3\right)}{2}-\frac{x\left(x-3\right)\left(2x+1\right)}{\sqrt{2x^3-x^2+x+1}+2x+1}=0\)

\(\Leftrightarrow x\left(x-3\right)\left(\frac{1}{2}-\frac{2x+1}{\sqrt{2x^3-x^2+x+1}+2x+1}\right)=0\)

Pt trong ngoặc vô nghiệm nốt nên 

\(\orbr{\begin{cases}x=0\\x-3=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=3\end{cases}}\)

e)Sửa đề \(\sqrt{\left(7+\sqrt{48}\right)^x}+\sqrt{\left(7-\sqrt{48}\right)^x}=14\)

\(\Leftrightarrow\left(7+\sqrt{48}\right)^{\frac{x}{2}}+\left(7-\sqrt{48}\right)^{\frac{x}{2}}=14\)

Dễ thấy: \(7+\sqrt{48}\) là số đối của \(7-\sqrt{48}\), nên

\(\left(7+\sqrt{48}\right)^{\frac{x}{2}}\left(7-\sqrt{48}\right)^{\frac{x}{2}}=1\)

Đặt \(\left(7-\sqrt{48}\right)^{\frac{x}{2}}=t\) thì ta dc pt

\(t+\frac{1}{t}=14\Rightarrow t^2-14t+1=0\)\(\Rightarrow t=7\pm4\sqrt{3}\)

Vì vậy \(t=7\pm4\sqrt{3}=\left(7-4\sqrt{3}\right)^{\pm1}=\left(7-4\sqrt{3}\right)^{\frac{x}{2}}\)

\(\Rightarrow\frac{x}{2}=\pm1\Rightarrow x=\pm2\)

g)\(\sqrt{17-x^2}=\left(3-\sqrt{x}\right)^2\)

Đk:\(0\le x\le\sqrt{17}\)

\(pt\Leftrightarrow\sqrt{17-x^2}-\left(-x+5\right)=\left(3-\sqrt{x}\right)^2-\left(-x+5\right)\)

\(\Leftrightarrow\frac{17-x^2-\left(x-5\right)^2}{\sqrt{17-x^2}-x+5}=x-6\sqrt{x}+9+x-5\)

\(\Leftrightarrow\frac{-2\left(x-1\right)\left(x-4\right)}{\sqrt{17-x^2}-x+5}-2\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)=0\)

\(\Leftrightarrow\frac{-2\left(x-1\right)\left(x-4\right)}{\sqrt{17-x^2}-x+5}-\frac{2\left(x-1\right)\left(x-4\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-4\right)\left(\frac{-2}{\sqrt{17-x^2}-x+5}-\frac{2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}\right)=0\)

Pt trong ngoặc <0 

Nên x-1=0 hoặc x-4=0 Hay x=1 hoặc x=4

1.

ĐKXĐ: \(x\ge-3\)

\(\Leftrightarrow\sqrt{\left(x+3\right)\left(x+7\right)}-3\sqrt{x+3}-2\left(\sqrt{x+7}-3\right)=0\)

\(\Leftrightarrow\sqrt{x+3}\left(\sqrt{x+7}-3\right)-2\left(\sqrt{x+7}-3\right)=0\)

\(\Leftrightarrow\left(\sqrt{x+3}-2\right)\left(\sqrt{x+7}-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+3}=2\\\sqrt{x+7}=3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

2.

\(\Leftrightarrow\sqrt{x}-3+\sqrt{x-5}-2+\sqrt{x+7}-4=0\)

\(\Leftrightarrow\frac{x-9}{\sqrt{x}+3}+\frac{x-9}{\sqrt{x-5}+2}+\frac{x-9}{\sqrt{x+7}+4}=0\)

\(\Leftrightarrow\left(x-9\right)\left(\frac{1}{\sqrt{x}+3}+\frac{1}{\sqrt{x-5}+2}+\frac{1}{\sqrt{x+7}+4}\right)=0\)

\(\Leftrightarrow x=9\)

3.

ĐKXĐ: \(\left[{}\begin{matrix}x>2\\x\le-1\end{matrix}\right.\)

- Với \(x>2\)

\(\Leftrightarrow\left(x-2\right)\left(x+1\right)+3\sqrt{\frac{\left(x-2\right)^2\left(x+1\right)}{x-2}}-10=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+1\right)+3\sqrt{\left(x-2\right)\left(x+1\right)}-10=0\)

Đặt \(\sqrt{\left(x-2\right)\left(x+1\right)}=t\ge0\)

\(\Rightarrow t^2+3t-10=0\Rightarrow\left[{}\begin{matrix}t=2\\t=-5\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{\left(x-2\right)\left(x+1\right)}=2\Leftrightarrow\left(x-2\right)\left(x+1\right)=4\Leftrightarrow...\)

- Với \(x\le-1\Rightarrow x-2< 0\Rightarrow x-2=-\sqrt{\left(x-2\right)^2}\)

Pt tương đương:

\(\left(x-2\right)\left(x+1\right)-3\sqrt{\frac{\left(x-2\right)^2\left(x+1\right)}{x-2}}-10=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+1\right)-3\sqrt{\left(x-2\right)\left(x+1\right)}-10=0\)

Đặt \(\sqrt{\left(x-2\right)\left(x+1\right)}=t\ge0\Rightarrow t^2-3t-10=0\Rightarrow t=5\)

\(\Rightarrow\left(x-2\right)\left(x+1\right)=25\Leftrightarrow...\)

4.

ĐKXĐ: \(x\ge4\)

Đặt \(\sqrt{x-4}=t\ge0\Rightarrow x=t^2+4\)

\(\Rightarrow3\left(t^2+4\right)+7t=14t-20\)

\(\Leftrightarrow3t^2-7t+34=0\)

Phương trình vô nghiệm

5.

ĐKXĐ: ...

- Với \(x=0\) ko phải nghiệm

- Với \(x\ne0\Rightarrow\sqrt{x+1}-1\ne0\) , nhân 2 vế của pt cho \(\sqrt{x+1}-1\) và rút gọn ta được:

\(\sqrt{x+1}+2x-5=\sqrt{x+1}-1\)

\(\Leftrightarrow2x=4\Rightarrow x=2\)

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

\(A=\left(3\cdot1+8\cdot1+2\right)^{2018}=13^{2018}\)