x/5= 2/3
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(Tớ sửa lại đề nhé.)
\(\dfrac{x-5}{x^2-9}-\dfrac{5}{3-x}=\dfrac{4}{x+3}\)
Điều kiện: \(x\ne\pm3\)
\(\Leftrightarrow\dfrac{x-5}{\left(x+3\right)\left(x-3\right)}+\dfrac{5\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}=\dfrac{4\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\)
\(\Rightarrow x-5+5x+15=4x-12\)
\(\Leftrightarrow x+5x-4x=-12-15+5\)
\(\Leftrightarrow2x=-22\)
\(\Leftrightarrow x=-11\)
[5(x + 2y)⁶ - 6(x + 2y)⁵] : [2(x + 2y)⁴]
= 5(x + 2y)⁶ : [2(x + 2y)⁴] - 6(x + 2y)⁵ : [2(x + 2y)⁴]
= 5(x + 2y)²/2 - 3(x + 2y)
\(\left(x-3\right)^2+\left(x+2\right)\left(5-x\right)\)
\(=x^2-6x+9+\left(5x-x^2+10-2x\right)\)
\(=x^2-6x+9+3x-x^2+10\)
\(=-3x+19\)
x(x-5).(x+5)-(x+2).(x^2-2x+4)=17
\(\Leftrightarrow x\left(x^2-25\right)-\left(x^3-2x^2+4x+2x^2-4x+8\right)=17\)
\(\Leftrightarrow x^3-25x-x^3+2x^2-4x-2x^2+4x-8=17\)
\(\Leftrightarrow-25x=17+8\)
\(\Leftrightarrow-25x=25\)
\(\Leftrightarrow x=-1\)
#)Giải :
\(x\left(x-5\right)\left(x+5\right)-\left(x+2\right)\left(x^2-2x+4\right)=17\)
\(\Rightarrow x\left(x^2-25\right)-\left(x^3-2x^2+4x+2x^2-4x+8\right)=17\)
\(\Rightarrow x^3-25-\left(x^3+8\right)=17\)
\(\Rightarrow x^3-25x-x^3-8=17\)
\(\Rightarrow-25x=25\Rightarrow x=-1\)
Vậy x = -1
\(\left(x+1\right)\left(x+3\right)-\left(x-2\right)\left(x+5\right)=2\)
\(\Leftrightarrow x^2+4x+3-x^2-3x+10=2\)
\(\Leftrightarrow x=-11\)
\(2\cdot11^x=\left(3^2+2\right)^3:\left(5^3-2^5:2^3\right)\)
\(\Leftrightarrow11^x\cdot2=1331:121\)
\(\Leftrightarrow11^x\cdot2=11\)
=> Phương trình vô nghiệm
\(\frac{x}{5}=\frac23\)
\(\Rightarrow x.3=5.2\)
\(x.3=10\)
\(x=10:3\)
\(x=\frac{10}{3}\)
\(\dfrac{x}{5}=\dfrac{2}{3}\)
=>\(x=\dfrac{2}{3}\cdot5\)
=>\(x=\dfrac{10}{3}\)