\(Đk:\) \(x\ne1,x\ne2,x\ne3\)

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

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

\(\Rightarrow x^2-3x+4x-12+x^2-2x+x-2=2x^2-4x+5x-10\)

\(\Rightarrow0x-14=x-10\)

\(\Rightarrow x=-4\left(tmđk\right)\)

\(a)C_6H_6 + 3Cl_2 \to C_6H_6Cl_6\\ n_{Cl_2}= \dfrac{2,24}{22,4} = 0,1(mol)\\ n_{C_6H_6} = \dfrac{1}{3}n_{Cl_2} = \dfrac{0,1}{3}(mol)\\ m_{C_6H_6} = \dfrac{0,1}{3}.78 = 26(gam)\\ b) n_{C_6H_6Cl_6} = n_{C_6H_6} = \dfrac{0,1}{3}(mol)\\ m_{C_6H_6Cl_6} = \dfrac{0,1}{3}.291 = 97(gam)\)

Hicc cảm ơn nhìu ạ❤️

um

Bruh.

Câu 4:

a: ΔOAB vuông tại A

=>\(\hat{AOB}+\hat{ABO}=90^0\)

=>\(\hat{ABO}=90^0-50^0=40^0\)

At//OB

=>\(\hat{tAB}=\hat{OBA}\) (hai góc so le trong)

=>\(\hat{tAB}=40^0\)
TA có: \(\hat{xAt}=\hat{xOB}\) (hai góc đồng vị, At//OB)

mà \(\hat{xOB}=50^0\)

nên \(\hat{xAt}=50^0\)

b: AH⊥Oy

Oy//At

Do đó: AH⊥ At

c: Xét ΔMHB có \(\hat{AMH}\) là góc ngoài tại đỉnh M

nên \(\hat{AMH}=\hat{MHB}+\hat{MBH}=\hat{MHB}+\hat{MAt}\)

Ta có: \(\frac{a+b}{b+c}=\frac79\)

=>\(\frac{a+b}{7}=\frac{b+c}{9}\)

\(\frac{c+a}{a+b}=\frac87\)

=>\(\frac{a+c}{8}=\frac{a+b}{7}\)

=>\(\frac{a+b}{7}=\frac{a+c}{8}=\frac{b+c}{9}\)

Đặt \(\frac{a+b}{7}=\frac{a+c}{8}=\frac{b+c}{9}=k\)

=>a+b=7k; a+c=8k; b+c=9k

=>a+c-a-b=8k-7k=k; c+b=9k; a+b=7k

=>c-b=k và c+b=9k và a+b=7k

=>c=(k+9k)/2=5k; b=9k-5k=4k; a=7k-4k=3k

\(P=\frac{3ab+2bc+ac}{a^2+2b^2+3c^2}\)

\(=\frac{3\cdot3k\cdot4k+2\cdot4k\cdot5k+3k\cdot5k}{\left(3k\right)^2+2\cdot\left(4k\right)^2+3\cdot\left(5k\right)^2}\)

\(=\frac{36k^2+40k^2+15k^2}{9k^2+32k^2+75k^2}=\frac{36+40+15}{9+32+75}=\frac{91}{116}\)