Yêu cầu?

a, \(\sqrt{\left(2x+3\right)^2}=x+1\)

\(\Leftrightarrow\left|2x+3\right|=x+1\)

TH1: \(\left\{{}\begin{matrix}2x+3=x+1\\2x+3\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\x\ge-\dfrac{3}{2}\end{matrix}\right.\Rightarrow\) vô nghiệm.

Vậy phương trình vô nghiệm.

TH2: \(\left\{{}\begin{matrix}-2x-3=x+1\\2x+3< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{4}{3}\\x< -\dfrac{3}{2}\end{matrix}\right.\Rightarrow\) vô nghiệm.

b, 

a, \(\sqrt{\left(2x-1\right)^2}=x+1\)

\(\Leftrightarrow\left|2x-1\right|=x+1\)

TH1: \(\left\{{}\begin{matrix}2x-1=x+1\\2x-1\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x\ge\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow x=2\)

TH2: \(\left\{{}\begin{matrix}-2x+1=x+1\\2x-1< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x< \dfrac{1}{2}\end{matrix}\right.\Leftrightarrow x=0\)

tìm x, biết

\(a,\dfrac{x+2\sqrt{x}-3}{\sqrt{x}-1}\)

\(\Leftrightarrow\dfrac{x+3\sqrt{x}-\sqrt{x}-3}{\sqrt{x}-1}\)

\(\Leftrightarrow\dfrac{\sqrt{x}.\left(\sqrt{x}+3\right)-\left(\sqrt{x}+3\right)}{\sqrt{x}-1}\)

\(\Leftrightarrow\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)

\(\Rightarrow\sqrt{x}+3\)

\(b,\dfrac{4y+3\sqrt{y}-7}{4\sqrt{y}+7}\)

\(\Leftrightarrow\dfrac{4y+7\sqrt{y}-4\sqrt{y}-7}{4\sqrt{y}+7}\)

\(\Leftrightarrow\dfrac{\sqrt{y}.\left(4\sqrt{y}\right)-\left(4\sqrt{y}+7\right)}{4\sqrt{y}+7}\)

\(\Leftrightarrow\dfrac{\left(4\sqrt{y}+7\right).\left(\sqrt{y}-1\right)}{4\sqrt{y}+7}\)

\(\Rightarrow\sqrt{y}-1\)

\(c,\dfrac{x\sqrt{y}-y\sqrt{x}}{\sqrt{x}-\sqrt{y}}\)

\(\Leftrightarrow\dfrac{\sqrt{xy}.\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}\)

\(\Rightarrow\sqrt{xy}\)

\(d,\dfrac{x-3\sqrt{x}-4}{x-\sqrt{x}-12}\)

\(\Leftrightarrow\dfrac{x+\sqrt{x}-4\sqrt{x}-4}{x+3\sqrt{x}-4\sqrt{x}-12}\)

\(\Leftrightarrow\dfrac{\sqrt{x}.\left(\sqrt{x}+1\right)-4\left(\sqrt{x}+1\right)}{\sqrt{x}.\left(x+3\right)-4\left(\sqrt{x}+3\right)}\)

\(\Leftrightarrow\dfrac{\left(\sqrt{x}+1\right).\left(\sqrt{x}-4\right)}{\left(\sqrt{x}+3\right).\left(\sqrt{x}-4\right)}\)

\(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)

\(\Rightarrow\dfrac{x-2\sqrt{x}-3}{x-9}\)

\(e,\dfrac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{1+\sqrt{4}}\)

\(\Leftrightarrow\dfrac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{1+2}\)

\(\Rightarrow\dfrac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{3}\)

\(f,\sqrt{8-2\sqrt{15}}+\sqrt{5}+\sqrt{3}\)

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

\(\Leftrightarrow\sqrt{5}-\sqrt{3}+\sqrt{5}+\sqrt{3}\)

\(\Rightarrow2\sqrt{5}-\sqrt{3}+\sqrt{3}\)

\(g,\sqrt{9-2\sqrt{4}}-\sqrt{9+2\sqrt{14}}\)

\(\Leftrightarrow\sqrt{9-2\times2}-\sqrt{\left(\sqrt{2}+\sqrt{7}\right)^2}\)

\(\Leftrightarrow\sqrt{9-4}-\left(\sqrt{2}+\sqrt{7}\right)\)

\(\Rightarrow\sqrt{5}-\sqrt{2}-\sqrt{7}\)

Câu 4: a) ĐK: \(x^2\ge9\Leftrightarrow\left[{}\begin{matrix}x\ge3\\x\le-3\end{matrix}\right.\)

b) ĐK: \(x^2-3x+2\ge0\Leftrightarrow\left[{}\begin{matrix}x\le1\\x\ge2\end{matrix}\right.\)

c) Đk: \(-3\le x< 5\)

d) x + 3 và 5 - x đồng dấu. Xét hai trường hợp:

\(\left\{{}\begin{matrix}x+3\ge0\\5-x>0\left(\text{do mẫu phải khác 0}\right)\end{matrix}\right.\Leftrightarrow-3\le x< 5\)

\(\left\{{}\begin{matrix}x+3< 0\\5-x< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< -3\\x>5\end{matrix}\right.\) do x ko thể đồng thời thỏa mãn cả hai nên loại.

Câu 1:

a) Đặt \(A=x+\sqrt{\left(x+2\right)^2}\cdot\left(x-2\right)\)

\(A=x+\left|x+2\right|\cdot\left(x-2\right)\)

+) Với \(x\ge-2\):

\(A=x+\left(x+2\right)\left(x-2\right)=x+x^2-4\)

+) Với \(x< -2\):

\(A=x-\left(x+2\right)\left(x-2\right)=x-x^2+4\)

b) \(B=\sqrt{m^2-6m+9-2m}\)

\(B=\sqrt{m^2-8m+9}\)

Bạn xem lại đề nhé :)

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

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

d) \(D=\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}\)

\(D=\sqrt{x-4+4\sqrt{x-4}+4}+\sqrt{x-4-4\sqrt{x-4}+4}\)

\(D=\sqrt{\left(\sqrt{x-4}+2\right)^2}+\sqrt{\left(\sqrt{x-4}-2\right)^2}\)

\(D=\sqrt{x-4}+2+\left|\sqrt{x-4}-2\right|\)

+) Xét \(x\ge8\):

\(D=\sqrt{x-4}+2+\sqrt{x-4}-2=2\sqrt{x-4}\)

+) Xét \(4< x< 8\):

\(D=\sqrt{x-4}+2+2-\sqrt{x-4}=4\)

Vậy....

Câu 2:

a) Ta có: \(\sqrt{5}< \sqrt{9}=3\)

b) \(2\sqrt{2}< \left(2+1\right)\sqrt{2}=3\sqrt{2}\)

c) \(-4\sqrt{5}>-4\sqrt{6}>-6\sqrt{6}\)

d) Xét hiệu: \(2\sqrt{3}-5-\sqrt{3}+4=\sqrt{3}-1>\sqrt{1}-1=0\)

Nên \(2\sqrt{3}-5>\sqrt{3}-4\)

e) Tương tự

chú ý\(x=\sqrt{x}^2\) tương tự với y , và các số tự nhiên dương

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

\(B=\frac{\left(2\sqrt{y}\right)^2+3\sqrt{y}-7}{4\sqrt{y}+7}=\frac{\left(\sqrt{y}-1\right)\left(4\sqrt{y}+7\right)}{4\sqrt{y}+7}=\sqrt{y}-1\)

\(C=\frac{\sqrt{x}^2\sqrt{y}-\sqrt{y}^2\sqrt{x}}{\sqrt{x}-\sqrt{y}}=\frac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}=\sqrt{xy}\)

\(D=\frac{\sqrt{x}^2-3\sqrt{x}-4}{\sqrt{x}^2-\sqrt{x}-12}=\frac{\left(\sqrt{x}-4\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+3\right)}=\frac{\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)}\)

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

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

CÂU CUỐI chưa làm đc

ý cuối cùng này :

\(D=\sqrt{13-4\sqrt{10}}+\sqrt{13+4\sqrt{10}}\)lấy bình phương 2 vế ta có

\(D^2=13-4\sqrt{10}+13+4\sqrt{10}+2\sqrt{13-4\sqrt{10}}\sqrt{13+4\sqrt{10}}\)

\(D^2=26+2\sqrt{13^2-16\sqrt{10}^2}\Leftrightarrow D^2=26+2\sqrt{9}\)

\(D^2=32\Leftrightarrow D=\sqrt{32}=4\sqrt{2}\)

Bài 4 :

\(a,\sqrt{x-1}=2\)

=> \(x-1=2^2=4\)

=>\(x=4+1=5\)

Vậy \(x\in\left\{5\right\}\)

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

=> \(x^2-3x+2=2\)

=> \(x^2-3x=2-2=0\)

=>\(x.\left(x-3\right)=0\)( phân tích đa thức thanh nhân tử )

=> \(\left[{}\begin{matrix}x=0\\x-3=0=>x=0+3=3\end{matrix}\right.\)

Vậy \(x\in\left\{0;3\right\}\)

MÌNH Biết vậy thôi ,

Bài 4 :

c) \(\sqrt{4x+1}=x+1\)ĐK : \(x\ge-1\)

\(\Leftrightarrow4x+1=\left(x+1\right)^2\)

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

\(\Leftrightarrow x^2-2x=0\)

\(\Leftrightarrow x\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)( thỏa )

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

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

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

\(\Leftrightarrow\left|\sqrt{x-1}+1\right|-\left|\sqrt{x-1}-1\right|=2\)

+) Xét \(x\ge2\)

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

\(\Leftrightarrow2=2\)( luôn đúng )

+) Xét \(1\le x< 2\):

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

\(\Leftrightarrow\sqrt{x-1}=1\)

\(\Leftrightarrow x-1=1\)

\(\Leftrightarrow x=2\)( loại )

Vậy \(x\ge2\)

a) Để \(\sqrt{\dfrac{x}{3}}\) có nghĩa thì \(\dfrac{x}{3}\ge0\Leftrightarrow x\ge0\)

b) Để \(\sqrt{-5x}\) có nghĩa thì \(-5x\ge0\Leftrightarrow x\le0\)

c) Để \(\sqrt{4-x}\) có nghĩa thì \(4-x\ge0\Leftrightarrow x\le4\)

d) Để \(\sqrt{3x+7}\) có nghĩa thì \(3x+7\ge0\Leftrightarrow x\ge-\dfrac{7}{3}\)

e) Để \(\sqrt{-3x+4}\) có nghĩa thì \(-3x+4\ge0\Leftrightarrow x\le\dfrac{4}{3}\)

f) Để \(\sqrt{\dfrac{1}{-1+x}}\) có nghĩa thì \(\left\{{}\begin{matrix}\dfrac{1}{-1+x}\ge0\\-1+x\ne0\end{matrix}\right.\)

\(\Leftrightarrow-1+x>0\Leftrightarrow x>1\)

g) Để \(\sqrt{1+x^2}\) có nghĩa thì \(1+x^2\ge0\left(đúng\forall x\right)\)

h) \(\sqrt{\dfrac{5}{x-2}}\) có nghĩ thì \(\left\{{}\begin{matrix}\dfrac{5}{x-2}\ge0\\x-2\ne0\end{matrix}\right.\)

\(\Leftrightarrow x-2>0\Leftrightarrow x>2\)

a. \(x\ge0\)

b. \(x< 0\)

c. \(x\le4\)

d. \(x\ge\dfrac{-7}{3}\)

e. \(x\le\dfrac{4}{3}\)

f. \(x>1\)

g. Mọi x

h. \(x>2\)

a.

ĐKXĐ: $x\geq 0; y\geq 1$

PT $\Leftrightarrow (x-4\sqrt{x}+4)+(y-1-6\sqrt{y-1}+9)=0$
$\Leftrightarrow (\sqrt{x}-2)^2+(\sqrt{y-1}-3)^2=0$
Vì $(\sqrt{x}-2)^2; (\sqrt{y-1}-3)^2\geq 0$ với mọi $x\geq 0; y\geq 1$ nên để tổng của chúng bằng $0$ thì:

$\sqrt{x}-2=\sqrt{y-1}-3=0$

$\Leftrightarrow x=4; y=10$

b.

ĐKXĐ: $x\geq -1; y\geq -2; z\geq -3$
PT $\Leftrightarrow x+y+z+35-4\sqrt{x+1}-6\sqrt{y+2}-8\sqrt{z+3}=0$

$\Leftrightarrow [(x+1)-4\sqrt{x+1}+4]+[(y+2)-6\sqrt{y+2}+9]+[(z+3)-8\sqrt{z+3}+16]=0$

$\Leftrightarrow (\sqrt{x+1}-2)^2+(\sqrt{y+2}-3)^2+(\sqrt{z+3}-4)^2=0$
$\Rightarrow \sqrt{x+1}-2=\sqrt{y+2}-3=\sqrt{z+3}-4=0$
$\Rightarrow x=3; y=7; z=13$

c.

ĐKXĐ: $x\geq \frac{-1}{8}$

PT $\Leftrightarrow 9x+17-6\sqrt{8x+1}-4\sqrt{x+3}=0$

$\Leftrightarrow [(8x+1)-6\sqrt{8x+1}+9]+[(x+3)-4\sqrt{x+3}+4]=0$

$\Leftrightarrow (\sqrt{8x+1}-3)^2+(\sqrt{x+3}-2)^2=0$

$\Rightarrow \sqrt{8x+1}-3=\sqrt{x+3}-2=0$

$\Rightarrow x=1$ (thỏa mãn đkxđ)

1.

ĐKXĐ: \(x\ge\frac{1}{2}\)

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

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

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

\(\)\(\Leftrightarrow x=1\)

2.

ĐKXĐ: ...

Đặt \(\left\{{}\begin{matrix}\sqrt{x^2+x+1}=a>0\\\sqrt{x^2-3x-1}=b\ge0\end{matrix}\right.\)

\(\Rightarrow a=b+\frac{1}{2}\left(a^2-b^2\right)\)

\(\Leftrightarrow\left(a-b\right)\left(a+b-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=b\left(1\right)\\a=2-b\left(2\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow x^2+x+1=x^2-3x-1\)

\(\Leftrightarrow x=\frac{1}{2}\)

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

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

\(\Rightarrow2\sqrt{x^2-3x-1}=1-2x\)

\(\Rightarrow4x^2-12x-4=4x^2-4x+1\)

\(\Rightarrow x=-\frac{5}{8}\)

Do các bước biến đổi ko tương đương nên cần thay nghiệm này vào pt ban đầu để kiểm tra (bạn tự kiểm tra)

3.

- Với \(x=\left\{16;17\right\}\) là 2 nghiệm của pt

- Với \(x< 16\):

\(\left\{{}\begin{matrix}\left|x-16\right|^4>0\\\left|x-17\right|>1\Rightarrow\left|x-17\right|^3>1\end{matrix}\right.\)

\(\Rightarrow\left|x-16\right|^4+\left|x-17\right|^3>1\)

Pt vô nghiệm

- Với \(x>17\Rightarrow\left\{{}\begin{matrix}\left|x-17\right|^3>0\\\left|x-16\right|>1\Rightarrow\left|x-16\right|^4>1\end{matrix}\right.\)

\(\Rightarrow\left|x-16\right|^4+\left|x-17\right|^3>1\)

Pt vô nghiệm

- Với \(16< x< 17\Rightarrow\left\{{}\begin{matrix}0< \left|x-16\right|< 1\\0< \left|17-x\right|< 1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left|x-16\right|^4< x-16\\\left|17-x\right|^3< 17-x\end{matrix}\right.\)

\(\Rightarrow\left|x-16\right|^4+\left|x-17\right|^3< x-16+17-x=1\) (vô nghiệm)

Vậy pt có đúng 2 nghiệm \(\left[{}\begin{matrix}x=16\\x=17\end{matrix}\right.\)