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Bài 2:
1: =>5x+1=6/7 hoặc 5x+1=-6/7
=>5x=-1/7 hoặc 5x=-13/7
=>x=-1/35 hoặc x=-13/35
2: =>x-1=4
=>x=5
3: =>3x-1=3
=>3x=4
=>x=4/3
4: \(\Leftrightarrow\dfrac{5}{x+3}=\dfrac{-5}{6}+\dfrac{1}{2}=\dfrac{-5+3}{6}=\dfrac{-2}{6}=\dfrac{-1}{3}\)
=>x+3=-15
=>x=-18
7: \(\Leftrightarrow2^{2x+1}+2^{2x+6}=264\)
=>2^2x+1*(1+2^5)=264
=>2^2x+1=8
=>2x+1=3
=>x=1
9: =>x^4=8x
=>x^4-8x=0
=>x=2
\(\dfrac{\left(13\dfrac{1}{4}-1\dfrac{5}{27}-10\dfrac{5}{6}\right).230\dfrac{1}{25}+46\dfrac{3}{4}}{\left(1\dfrac{3}{7}+\dfrac{10}{3}\right):\left(12\dfrac{1}{3}-14\dfrac{2}{7}\right)}\)
\(=\dfrac{1\dfrac{25}{108}.230\dfrac{1}{25}+46\dfrac{3}{4}}{4\dfrac{16}{21}:\left(-1\dfrac{20}{21}\right)}=\dfrac{330\dfrac{1}{25}}{-2\dfrac{18}{41}}=-135,3164\)
Bài 7:
x/1=z/2 nên x/6=z/12
=>x/6=y/9=z/12
=>x/2=y/3=z/4
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y+z}{2+3+4}=\dfrac{27}{9}=3\)
=>x=6; y=9; z=12
a, \(\dfrac{13}{32}+\dfrac{8}{24}+\dfrac{19}{32}+\dfrac{2}{3}\)
\(=\left(\dfrac{13}{32}+\dfrac{19}{32}\right)+\left(\dfrac{1}{3}+\dfrac{2}{3}\right)\)
\(=\dfrac{32}{32}+\dfrac{3}{3}=1+1=2\)
b, \(\dfrac{3}{4}.36\dfrac{1}{5}-\dfrac{3}{4}.2\dfrac{1}{5}\)
\(=\dfrac{3}{4}.\left(36\dfrac{1}{5}-2\dfrac{1}{5}\right)\)
\(=\dfrac{3}{4}.\left[\left(36-2\right)+\left(\dfrac{1}{5}-\dfrac{1}{5}\right)\right]\)
\(=\dfrac{3}{4}.34=\dfrac{102}{4}=26\)
Bài 2:
a: x=27/10:9/5=27/10*5/9=135/90=3/2
b: =>|x|=1,75
=>x=1,75 hoặc x=-1,75
c: =>\(2-x=\sqrt[3]{25}\)
hay \(x=2-\sqrt[3]{25}\)
d: =>3^x-1*6=162
=>3^x-1=27
=>x-1=3
=>x=4
1.Tính
a.\(\dfrac{7}{23}\left[(-\dfrac{8}{6})-\dfrac{45}{18}\right]=\dfrac{7}{23}.-\dfrac{12}{6}=-\dfrac{7}{6}\)
b.\(\dfrac{1}{5}\div\dfrac{1}{10}-\dfrac{1}{3}(\dfrac{6}{5}-\dfrac{9}{4})=2-(-\dfrac{7}{20})=\dfrac{47}{20}\)
c.\(\dfrac{3}{5}.(-\dfrac{8}{3})-\dfrac{3}{5}\div(-6)=-\dfrac{3}{2}\)
d.\(\dfrac{1}{2}.(\dfrac{4}{3}+\dfrac{2}{5})-\dfrac{3}{4}.(\dfrac{8}{9}+\dfrac{16}{3})=-\dfrac{19}{5}\)
e.\(\dfrac{6}{7}\div(\dfrac{3}{26}-\dfrac{3}{13})+\dfrac{6}{7}.(\dfrac{1}{10}-\dfrac{8}{5})=-\dfrac{61}{7}\)
Bài 2
a.\(1^2_5x+\dfrac{3}{7}=\dfrac{4}{5}\)
\(x=\dfrac{13}{49}\)
b.\(\left|x-1,5\right|=2\)
Xảy ra 2 trường hợp
TH1
\(x-1,5=2\)
\(x=3,5\)
TH2
\(x-1,5=-2\)
\(x=-0,5\)
Vậy \(x=3,5\) hoặc \(x=-0,5\) .
Ngại làm quá trời ơi,lần sau bn tách ra nhá làm vậy mỏi tay quá.
Bài 1:
a)
\(|x+\frac{4}{15}|-|-3,75|=-|-2,15|\)
\(\Leftrightarrow |x+\frac{4}{15}|-3,75=-2,15\)
\(\Leftrightarrow |x+\frac{4}{15}|=-2,15+3,75=\frac{8}{5}\)
\(\Rightarrow \left[\begin{matrix} x+\frac{4}{15}=\frac{8}{5}\\ x+\frac{4}{15}=-\frac{8}{5}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{4}{3}\\ x=\frac{-28}{15}\end{matrix}\right.\)
b )
\(|\frac{5}{3}x|=|-\frac{1}{6}|=\frac{1}{6}\)
\(\Rightarrow \left[\begin{matrix} \frac{5}{3}x=\frac{1}{6}\\ \frac{5}{3}x=-\frac{1}{6}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{1}{10}\\ x=-\frac{1}{10}\end{matrix}\right.\)
c)
\(|\frac{3}{4}x-\frac{3}{4}|-\frac{3}{4}=|-\frac{3}{4}|=\frac{3}{4}\)
\(\Leftrightarrow |\frac{3}{4}x-\frac{3}{4}|=\frac{3}{2}\)
\(\Rightarrow \left[\begin{matrix} \frac{3}{4}x-\frac{3}{4}=\frac{3}{2}\\ \frac{3}{4}x-\frac{3}{4}=-\frac{3}{2}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=3\\ x=-1\end{matrix}\right.\)
Bài 3:
a) Ta thấy:
\(|x+\frac{15}{19}|\geq 0, \forall x\Rightarrow A\ge 0-1=-1\)
Vậy GTNN của $A$ là $-1$ khi \(x+\frac{15}{19}=0\Leftrightarrow x=-\frac{15}{19}\)
b)Vì \(|x-\frac{4}{7}|\geq 0, \forall x\Rightarrow B\geq \frac{1}{2}+0=\frac{1}{2}\)
Vậy GTNN của $B$ là $\frac{1}{2}$ khi \(x-\frac{4}{7}=0\Leftrightarrow x=\frac{4}{7}\)
a: \(\left|x\right|=3+\dfrac{1}{5}=\dfrac{16}{5}\)
mà x<0
nên x=-16/5
b: \(\left|x\right|=-2.1\)
nên \(x\in\varnothing\)
c: \(\left|x-3.5\right|=5\)
=>x-3,5=5 hoặc x-3,5=-5
=>x=8,5 hoặc x=-1,5
d: \(\left|x+\dfrac{3}{4}\right|-\dfrac{1}{2}=0\)
=>|x+3/4|=1/2
=>x+3/4=1/2 hoặc x+3/4=-1/2
=>x=-1/4 hoặc x=-5/4
đề bài yêu cầu gì vậy em
tìm x biết :