tìm x biết
3 x+1+5.3x+2=144
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(3\left(x-2\right)+4\left(x-5\right)=23\)
\(\Rightarrow3x-6+4x-20-23=0\)
\(\Rightarrow7x-49=0\)
\(\Rightarrow x=7\)
3(x-2)+4(x-5)=23
<=>3x-6+4x-20=23
<=>7x-26=23
<=>7x=49
<=>x=7
Vậy x=7
Bài 1:
a) Ta có: \(\dfrac{17}{6}-x\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)
\(\Leftrightarrow\dfrac{17}{6}-x^2+\dfrac{7}{6}x-\dfrac{7}{4}=0\)
\(\Leftrightarrow-x^2+\dfrac{7}{6}x+\dfrac{13}{12}=0\)
\(\Leftrightarrow-12x^2+14x+13=0\)
\(\Delta=14^2-4\cdot\left(-12\right)\cdot13=196+624=820\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{14-2\sqrt{205}}{-24}=\dfrac{-7+\sqrt{205}}{12}\\x_2=\dfrac{14+2\sqrt{2015}}{-24}=\dfrac{-7-\sqrt{205}}{12}\end{matrix}\right.\)
b) Ta có: \(\dfrac{3}{35}-\left(\dfrac{3}{5}-x\right)=\dfrac{2}{7}\)
\(\Leftrightarrow\dfrac{3}{5}-x=\dfrac{3}{35}-\dfrac{10}{35}=\dfrac{-7}{35}=\dfrac{-1}{5}\)
hay \(x=\dfrac{3}{5}-\dfrac{-1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)
\(B=5\cdot\left(2x-1\right)^2+4\cdot\left(x-1\right)\cdot\left(x+3\right)-2\cdot\left(5\cdot3x\right)\)
\(=5\left(4x^2-4x+1\right)+\left(4x-4\right)\cdot\left(x+3\right)-2\cdot15x\)
\(=20x^2-20x+5+4x^2+12x-4x-12-2\cdot15x\)
\(=20x^2-20x+5+4x^2+12x-4x+12-30x\)
\(=24x^2-42x-7\)
(B=5left(2x-1 ight)^2+4left(x-1 ight)left(x+3 ight)-2left(5.3x ight)=5left(4x^2-4x+1 ight)+4left(x^2-4x+3 ight)-30x=20x^2-20x+5+4x^2-16x+12-30x=24x^2-66x+17)
\(a,5^x+5^{x+2}=650\\ \Rightarrow a,5^x+5^x.25=650\\ \Rightarrow26.5^x=650\\ \Rightarrow5^x=25\\ \Rightarrow5^x=5^2\\ \Rightarrow x=2\)
\(b,3^{x.1}+5.3^{x.1}=162\\ \Rightarrow3^x+5.3^x=162\\ \Rightarrow6.3^x=162\\ \Rightarrow3^x=27\\ \Rightarrow3^x=3^3\\ \Rightarrow x=3\)
Trả lời:
\(3x+1+5\times3x+2=144\)
\(\Leftrightarrow3x+1+15x+2=144\)
\(\Leftrightarrow18x=141\)
\(\Leftrightarrow x=\frac{47}{8}\)
Vậy \(x=\frac{47}{8}\)