a) \(\dfrac{3x-1}{x-1}-\dfrac{2x+5}{x+3}=1-\dfrac{4}{\left(x-1\right)\left(x+3\right)}\)

ĐKXĐ \(x-1\ne0\) hoặc \(x+3\ne0\)

\(\Rightarrow x\ne1\) và \(x\ne-3\)

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

\(\Leftrightarrow\left(3x-1\right)\left(x+3\right)-\left(2x+5\right)\left(x-1\right)=\left(x-1\right)\left(x+3\right)-4\)

\(\Leftrightarrow3x^2+9x-x-3-\left(2x^2-2x+5x-5\right)=x^2+3x-x-3-4\)

\(\Leftrightarrow3x^2+9x-x-3-2x^2+2x-5x+5=x^2+3x-x-3-4\)

\(\Leftrightarrow9x-x+2x-5x-3x+x=3-5-3-4\)

\(\Leftrightarrow3x=-9\)

\(\Leftrightarrow x=-3\) (không thỏa ĐK)

Vậy PTVN

b) \(\dfrac{13}{\left(x-3\right)\left(2x+7\right)}+\dfrac{1}{2x+7}=\dfrac{6}{\left(x-3\right)\left(x+3\right)}\)

ĐKXĐ: \(x-3\ne0\Rightarrow x\ne3\)

\(x+3\ne0\Rightarrow x\ne-3\)

\(2x+7\ne0\Rightarrow2x\ne-7\Rightarrow x\ne\dfrac{-7}{2}\)

\(\dfrac{13\left(x+3\right)}{\left(x-3\right)\left(x+3\right)\left(2x+7\right)}+\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)\left(2x+7\right)}=\dfrac{6\left(2x+7\right)}{\left(x-3\right)\left(x+3\right)\left(2x+7\right)}\)

\(\Leftrightarrow13\left(x+3\right)+\left(x-3\right)\left(x+3\right)=6\left(2x+7\right)\)

\(\Leftrightarrow13x+39+x^2+3x-3x-9=12x+42\)

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

\(\Leftrightarrow x^2-3x+4x-12=0\)

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

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

\(\left\{{}\begin{matrix}x-3=0\\x+4=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=3\left(KTĐK\right)\\x=-4\left(TĐK\right)\end{matrix}\right.\)

Vậy S={-4}

Theo đề bài, ta có:

\(\dfrac{\widehat{A}}{1}=\dfrac{\widehat{B}}{2}=\dfrac{\widehat{C}}{3}=\dfrac{\widehat{D}}{4}\)

Áp dụng tính chất của dãy tỉ số bằng nhau:

\(\dfrac{\widehat{A}}{1}=\dfrac{\widehat{B}}{2}=\dfrac{\widehat{C}}{3}=\dfrac{\widehat{D}}{4}\)=\(\dfrac{\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}}{1+2+3+4}=\dfrac{360^o}{10}=36^o\)

Ta có: \(\dfrac{\widehat{A}}{1}=36^o\Rightarrow\widehat{A}=36^o\)

\(\dfrac{\widehat{B}}{2}=36^o\Rightarrow\widehat{B}=72^o\)

\(\dfrac{\widehat{C}}{3}=36^o\Rightarrow\widehat{C}=108^o\)

\(\dfrac{\widehat{D}}{4}=36^o\Rightarrow\widehat{D}=144^o\)

a) \(\sqrt{2x+7}\)

Để \(\sqrt{2x+7}\) có nghĩa\(\Leftrightarrow\)2x+7\(\ge\)0

\(\Leftrightarrow\)2x\(\ge\)-7

\(\Leftrightarrow\)x\(\ge\)\(\dfrac{-7}{2}\)

b) \(\sqrt{-3x+4}\)

Để \(\sqrt{-3x+4}\) có nghĩa \(\Leftrightarrow\)-3x+4\(\ge\)0

\(\Leftrightarrow\)-3x\(\ge\)-4

\(\Leftrightarrow\)x\(\le\)\(\dfrac{4}{3}\)

c)\(\sqrt{\dfrac{1}{-1+x}}\)

Để \(\sqrt{\dfrac{1}{-1+x}}\) có nghĩa \(\Leftrightarrow\)\(\dfrac{1}{-1+x}\ge0\)

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

\(\Leftrightarrow\)x>1

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

Ta có x²+1\(\ge\)1>0;\(\forall\)x\(\in R\)

Vậy x\(\in R\)

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

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

c) \(x^2+2\sqrt{3}x+3=x^2+2.x.\sqrt{3}+\left(\sqrt{3}\right)^2=\left(x+\sqrt{3}\right)^2\)

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

a) \(x^2-5=0\)

\(x^2=5\Leftrightarrow x=-\sqrt{5}\) hoặc \(x=\sqrt{5}\)

Vậy S={\(-\sqrt{5}\);\(\sqrt{5}\)}

b) \(x^2-2.\sqrt{11}x+11=0\)

\(x^2-2.x.\sqrt{11}+\left(\sqrt{11}\right)^2=0\)

\(\left(x-\sqrt{11}\right)^2=0\)

\(x-\sqrt{11}=0\)

\(x=\sqrt{11}\)

Vậy S={\(\sqrt{11}\)}

\(\)

a) \(x^2-7=x^2-\left(\sqrt{7}\right)^2=\left(x-\sqrt{7}\right)\left(x+\sqrt{7}\right)\)

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

c) \(x^2+2\sqrt{13}x+13=x^2+2.x.\sqrt{13}+\left(\sqrt{13}\right)^2=\left(x+\sqrt{13}\right)^2\)

a) \(\sqrt{16}.\sqrt{25}+\sqrt{196}\div\sqrt{49}\)

\(=4.5+14:7\)

\(=20+2=22\)

b) \(36:\sqrt{2.3^2.18}-\sqrt{169}\)

\(=36:18-13=-11\)

c) \(\sqrt{\sqrt{81}}=\sqrt{9}=3\)

d) \(\sqrt{3^2+4^2}=\sqrt{25}=5\)