tìm x biết
3 phần x+1=1 phần 3
giup em vs
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\(=>x\left(\dfrac{1}{2}+\dfrac{2}{3}\right)=\dfrac{10}{3}+1=\dfrac{13}{3}\)
\(=>x=\dfrac{13}{3}:\left(\dfrac{1}{2}+\dfrac{2}{3}\right)=\dfrac{13}{3}:\dfrac{7}{6}=\dfrac{26}{7}\)
5/6x - 1 = 10/3
5/6x = 10/3 + 1 = 13/3
X = 13/3 : 5/6 = 26/5
\(\dfrac{5}{7}\times x+\dfrac{2}{5}=\dfrac{2}{3}\\ \dfrac{5}{7}\times x=\dfrac{2}{3}-\dfrac{2}{5}=\dfrac{10}{15}-\dfrac{6}{15}=\dfrac{4}{15}\\ x=\dfrac{4}{15}:\dfrac{5}{7}=\dfrac{4\times7}{15\times5}=\dfrac{28}{75}\)
Vậy `x=28/75`
`HaNa☘D`
\(\dfrac{5}{7}\times x+\dfrac{2}{5}=\dfrac{2}{3}\)
\(\dfrac{5}{7}\times x=\dfrac{2}{3}-\dfrac{2}{5}\)
\(\dfrac{5}{7}\times x=\dfrac{10}{15}-\dfrac{6}{15}\)
\(\dfrac{5}{7}\times x=\dfrac{4}{15}\)
\(x=\dfrac{4}{15}:\dfrac{5}{7}\)
\(x=\dfrac{4}{15}\times\dfrac{7}{5}\)
\(x=\dfrac{28}{75}\)
#Toru
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 \(x=\dfrac{3}{5}-\dfrac{-1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)
`1/2x^2(-2x^2y^2z).(-1/3)x^2y^3`
`=1/2 .(-2).(-1/3)x^{2+2+2}.y^{2+3}.z`
`=1/3x^6y^5z`
Ta có: \(\dfrac{1}{2}x^2\cdot\left(-2x^2y^2z\right)\cdot\left(-\dfrac{1}{3}\right)x^2y^3\)
\(=\left(\dfrac{1}{2}\cdot2\cdot\dfrac{1}{3}\right)\cdot\left(x^2\cdot x^2\cdot x^2\right)\cdot\left(y^2\cdot y^3\right)\cdot z\)
\(=\dfrac{1}{3}x^6y^5z\)
\(3\left(x-2\right)+4\left(x-5\right)=23\)
\(\Rightarrow3x-6+4x-20-23=0\)
\(\Rightarrow7x-49=0\)
\(\Rightarrow x=7\)
3(x-2)+4(x-5)=23
<=>3x-6+4x-20=23
<=>7x-26=23
<=>7x=49
<=>x=7
Vậy x=7
#)Giải :
\(\left(\left|x\right|-\frac{1}{8}\right)\left(-\frac{1}{8}\right)^5=\left(-\frac{1}{8}\right)^7\)
\(\Leftrightarrow\left(\left|x\right|-\frac{1}{8}\right)=\left(-\frac{1}{8}\right)^2\)
\(\Leftrightarrow\left(\left|x\right|-\frac{1}{8}\right)=\frac{1}{64}\)
\(\Leftrightarrow\left|x\right|=\frac{9}{64}\)
\(\Leftrightarrow\hept{\begin{cases}x>0\\x< 0\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{9}{64}\\x=-\frac{9}{64}\end{cases}}}\)