\(\frac{x}{30+3}+\frac{2}{3}=\frac{x}{30-3}\)

\(\Rightarrow\frac{x}{33}+\frac{2}{3}=\frac{x}{27}\)

\(\Rightarrow\frac{x}{33}=\frac{x}{27}-\frac{18}{27}\)

\(\Rightarrow\frac{x}{33}=\frac{x-18}{27}\)

\(\Rightarrow33\left(x-18\right)=27x\)

\(\Rightarrow33x-594=27x\)

\(\Rightarrow594=6x\)

\(\Rightarrow x=99\)

Ta có:

\(\frac{x}{30+3}+\frac{2}{3}=\frac{x}{30-3}\Rightarrow\frac{x}{33}+\frac{2}{3}=\frac{x}{27}\Rightarrow\frac{9x}{297}+\frac{198}{297}=\frac{11x}{297}\Rightarrow9x+198=11x\)

\(\Rightarrow11x-9x=198\Rightarrow2x=198\Rightarrow x=198:2\Rightarrow x=99\)

Vậy x = 99

Ta có : \(\dfrac{1}{3}\left(x-2\right)+\dfrac{1}{4}x\left(x-2\right)-\dfrac{1}{5}\left(x-2\right)=\dfrac{23}{30}\)

\(\Rightarrow20\left(x-2\right)+15x\left(x-2\right)-12\left(x-2\right)=46\)

\(\Leftrightarrow15x^2-30x+8x-16-46=0\)

\(\Leftrightarrow15x^2-22x-62=0\)

( Đến đây ra vô tỉ luôn : vvvv ; không biết đề này đúng chưa :vvv0

Bạn tham khảo ô này

để soạn thảo câu hỏi chính xác nha :vvvv

a) \({3^{x + 2}} = \sqrt[3]{9} \Leftrightarrow {3^{x + 2}} = {9^{\frac{1}{3}}} \Leftrightarrow {3^{x + 2}} = {\left( {{3^2}} \right)^{\frac{1}{3}}} \Leftrightarrow {3^{x + 2}} = {3^{\frac{2}{3}}} \Leftrightarrow x + 2 = \frac{2}{3} \Leftrightarrow x =  - \frac{4}{3}\)

b) \({2.10^{2{\rm{x}}}} = 30 \Leftrightarrow {10^{2{\rm{x}}}} = 15 \Leftrightarrow 2{\rm{x}} = \log 15 \Leftrightarrow x = \frac{1}{2}\log 15\)

c) \({4^{2{\rm{x}}}} = {8^{2{\rm{x}} - 1}} \Leftrightarrow {\left( {{2^2}} \right)^{2{\rm{x}}}} = {\left( {{2^3}} \right)^{2{\rm{x}} - 1}} \Leftrightarrow {2^{4{\rm{x}}}} = {2^{6{\rm{x}} - 3}} \Leftrightarrow 4{\rm{x}} = 6{\rm{x}} - 3 \Leftrightarrow  - 2{\rm{x}} =  - 3 \Leftrightarrow x = \frac{3}{2}\).

=>x/30-x/40=3/4

=>x/120=3/4

=>x=90

\(\dfrac{x}{30}=\dfrac{x}{40}+\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{4x}{120}=\dfrac{3x}{120}+\dfrac{90}{120}\)

\(\Leftrightarrow4x=3x+90\)

\(\Leftrightarrow4x-3x-90=0\)

\(\Leftrightarrow x-90=0\)

\(\Leftrightarrow x=90\)

\(\text{Vậy phương trình có tập nghiệm là }S=\left\{90\right\}\)

\(7x+6\sqrt{x+5}=x^2+30\left(đk:x\ge-5\right)\)

\(\Leftrightarrow6\sqrt{x+5}=x^2-7x+30\)

Ta thấy 2 vế đều dương nên bình phương lên ta được:

\(36x+180=x^4+49x^2+900-14x^3+60x^2-420x\)

\(\Leftrightarrow x^4-14x^3+109x^2-456x+720=0\)

\(\Leftrightarrow x^3\left(x-4\right)-10x^2\left(x-4\right)+69x\left(x-4\right)-180\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x^3-10x^2+69x-180\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left[x^2\left(x-4\right)-6x\left(x-4\right)+45\left(x-4\right)\right]=0\)

\(\Leftrightarrow\left(x-4\right)^2\left(x^2-6x+45\right)=0\)

\(\Leftrightarrow x=4\left(tm\right)\) (do \(x^2-6x+45=\left(x^2-6x+9\right)+36=\left(x-3\right)^2+36\ge36>0\))

a) \(5x - 30 = 0\)

\(5x = 0 + 30\)     

\(5x = 30\)

\(x = 30:5\)

\(x = 6\)      

Vậy phương trình có nghiệm \(x = 6\).

b) \(4 - 3x = 11\)

\( - 3x = 11 - 4\)

\( - 3x =  7\)

\(x = \left( { 7} \right):\left( { - 3} \right)\)

\(x = \dfrac{-7}{3}\)

Vậy phương trình có nghiệm \(x = \dfrac{7}{3}\).

c) \(3x + x + 20 = 0\)               

\(4x + 20 = 0\)

\(4x = 0 - 20\)

\(4x =  - 20\)

\(x = \left( { - 20} \right):4\)

\(x =  - 5\)   

Vậy phương trình có nghiệm \(x =  - 5\).

d) \(\dfrac{1}{3}x + \dfrac{1}{2} = x + 2\)

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

\(\dfrac{{ - 2}}{3}x = \dfrac{3}{2}\)

\(x = \dfrac{3}{2}:\left( {\dfrac{{ - 2}}{3}} \right)\)

\(x = \dfrac{{ - 9}}{4}\)

Vậy phương trình có nghiệm \(x = \dfrac{{ - 9}}{4}\).