a)\(\left(2x-3\right)\left(x+1\right)< 0\)

\(\Leftrightarrow\begin{cases}2x-3>0\\x+1< 0\end{cases}\)  hoặc \(\begin{cases}2x-3< 0\\x+1>0\end{cases}\)

\(\Leftrightarrow\begin{cases}x>\frac{3}{2}\\x< -1\end{cases}\) (loại)  hoặc \(\begin{cases}x< \frac{3}{2}\\x>-1\end{cases}\)

\(\Leftrightarrow-1< x< \frac{3}{2}\)

b) \(\left(x-\frac{1}{2}\right)\left(x+3\right)>0\)

\(\Leftrightarrow\begin{cases}x-\frac{1}{2}>0\\x+3>0\end{cases}\) hoặc \(\begin{cases}x-\frac{1}{2}< 0\\x+3< 0\end{cases}\)

\(\Leftrightarrow\begin{cases}x>\frac{1}{2}\\x>-3\end{cases}\) hoặc \(\begin{cases}x< \frac{1}{2}\\x< -3\end{cases}\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x>\frac{1}{2}\\x< -3\end{array}\right.\)

c) Sai đề phải là \(\frac{x}{\left(x+3\right)\left(x+7\right)}\)

Có: \(\frac{3}{\left(x+3\right)\left(x+5\right)}+\frac{5}{\left(x+5\right)\left(x+10\right)}+\frac{7}{\left(x+10\right)\left(x+17\right)}=\frac{x}{\left(x+3\right)\left(x+17\right)}\)

\(\Leftrightarrow\)\(\frac{1}{x+3}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+7}=\frac{x}{\left(x+3\right)\left(x+7\right)}\)

\(\Leftrightarrow\)\(\frac{1}{x+3}-\frac{1}{x+7}=\frac{x}{\left(x+3\right)\left(x+7\right)}\)

\(\Leftrightarrow\)\(\frac{4}{\left(x+3\right)\left(x+7\right)}=\frac{x}{\left(x+3\right)\left(x+7\right)}\)

\(\Leftrightarrow x=4\)

đề câu c sai rầu kìa má

a,\(\sqrt{x^2}=5\Rightarrow x=5\)

b,\(\sqrt{x}+5=7\Rightarrow\sqrt{x}=2\Rightarrow x=4\)

f,\(\frac{\sqrt{x-5}}{\sqrt{x-4}}=1\Rightarrow\sqrt{x-5}=\sqrt{x-4}\Rightarrow\left(\sqrt{x-5}\right)^2=\left(\sqrt{x-4}\right)^2\Rightarrow x-5=x-4\)

\(\Rightarrow x-x=5-4\Rightarrow0x=1\)(vô lý)  => x không tồn tại

Vì \(\left|x+\frac{3}{4}\right|\ge0;\left|y-\frac{1}{5}\right|\ge0;\left|x+y+z\right|\ge0\) với mọi x; y , z

 nên để \(\left|x+\frac{3}{4}\right|+\left|y-\frac{1}{5}\right|+\left|x+y+z\right|=0\)

thì \(\left|x+\frac{3}{4}\right|=\left|y-\frac{1}{5}\right|=\left|x+y+z\right|=0\)

=> \(x+\frac{3}{4}=0;y-\frac{1}{5}=0;x+y+z=0\)

+) x + 3/4 = 0 => x = -3/4

+) y - 1/5 = 0 => y =1/5

+) x + y + z = 0 => z = - x - y = 3/4 - 1/5 = 11/20

Từng cái trị tuyệt đối phải bằng 0 (vì GTTĐ luôn lớn hơn hoặc bằng 0 và tổng đó lại = 0)

1) x+3/4 = 0 => x = -3/4

2) y- 1/5 = 0 => y = 1/5

3) x+y+z=0 => -3/4 + 1/5 +z = 0 => z = 11/20

Vậy (x,y,z) = (-3/4;1/5;11/20)

a/ ĐKXĐ : \(x\ge0;x\ne1\)

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

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

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

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

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

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

\(=-\sqrt{x}\left(x-1\right)\)

Vậy...

b/ Ta có :

\(P>0\)

\(\Leftrightarrow-\sqrt{x}\left(x-1\right)>0\)

\(\Leftrightarrow\sqrt{x}\left(x-1\right)< 0\)

Mà \(\sqrt{x}\ge0\)

\(\Leftrightarrow x-1< 0\Leftrightarrow x< 1\)

Kết hợp ĐKXĐ

Vậy \(0< x< 1\) thì P > 0

c/ Ta có :

\(x=7-4\sqrt{3}=\left(2-\sqrt{3}\right)^2\) thỏa mãn \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

\(\Leftrightarrow\sqrt{x}=\left|2-\sqrt{3}\right|=2-\sqrt{3}\)

Thay vào P rồi bạn tự tính ra nhé :>

a)\(5^x.\left(5^3\right)^2=625\)

\(5^x.5^6=5^4\)

\(5^x=5^{-2}\)

\(x=-2\)

b)\(27< 81^3:3^x< 243\)

\(3^3< \left(3^4\right)^3:3^x< 3^5\)

\(3^3< 3^{12}:3^x< 3^5\)

\(3^{12}:3^x=3^4\)

\(3^x=3^3\)

\(x=3\)

c)\(\left(5x+1\right)^2=\frac{36}{49}\) 

\(\left(5x+1\right)^2=\left(\frac{6}{7}\right)^2\)

\(5x+1=\frac{6}{7}\)

\(5x=\frac{-1}{7}\)

\(x=\frac{-1}{35}\)

d)\(\left(x-\frac{2}{9}\right)^3=\left(\frac{2}{3}\right)^6\)

\(\left(x-\frac{2}{9}\right)^3=\left[\left(\frac{2}{3}\right)^2\right]^3\)

\(x-\frac{2}{9}=\frac{4}{9}\)

\(x=\frac{6}{9}=\frac{2}{3}\)

\(5^x.\left(5^3\right)^2=625\)

\(\Rightarrow5^x.5^6=5^4\)

\(\Rightarrow5^{x+6}=5^4\Rightarrow x+6=4\Rightarrow x=-2\)

Đề sai rồi bạn : Phải là :

 \(5^x:\left(5^3\right)^2=625\)

\(\Rightarrow5^x:5^6=5^4\)

\(\Rightarrow5^{x-6}=5^4\)

\(\Rightarrow x-6=4\Rightarrow x=10\)

Nhứng nếu đề đúng thì bạn có thể lấy KQ trên

1. Tìm x, biết :

a. ( x - \(\frac{3}{4}\)) \(^2\)= 0

=> x - \(\frac{3}{4}\)= 0

=> x = 0 + \(\frac{3}{4}\)

=> x = \(\frac{3}{4}\)

b. ( x + \(\frac{1}{2}\)) \(^2\)= \(\frac{9}{64}\)

=> ( x + \(\frac{1}{2}\)) \(^2\)= ( \(\frac{3}{8}\)) \(^2\)

=> x + \(\frac{1}{2}\)= \(\frac{3}{8}\)

=> x = \(\frac{3}{8}\)- \(\frac{1}{2}\)

=> x = \(\frac{-1}{8}\)

c.  \(\frac{\left(-2\right)^x}{16}=-8\)

=> \(\frac{\left(-2\right)^x}{16}=\frac{-8}{1}=\frac{-128}{16}\)

=> ( -2)\(^x\)= -128

=> ( -2 ) \(^x\)= ( -2) \(^7\)

=> x = 7