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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
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
\(xy-3x-y=6\)
\(=>xy+3x-y-3=6-3\)
\(=>x\left(y+3\right)-\left(y+3\right)=3\)
\(=>\left(y+3\right)\left(x-1\right)=3\)
| y+3 | -1 | 3 | 1 | -3 | |
| x-1 | -3 | 1 | 3 | -1 |
| y+3 | -1 | 3 | -3 | 1 |
| y | -4 | -1 | -7 | -3 |
| x-1 | -3 | 1 | 3 | -1 |
| x | -2 | 2 | 4 | 0 |
Bài 1:
a: \(=17+\dfrac{2}{31}-\dfrac{15}{17}-6-\dfrac{2}{31}=11-\dfrac{15}{17}=\dfrac{172}{17}\)
b: \(=31+\dfrac{6}{13}+5+\dfrac{9}{41}-36-\dfrac{9}{41}-36-\dfrac{6}{13}\)
=36
c: \(=27+\dfrac{51}{59}-7-\dfrac{51}{59}+\dfrac{1}{3}=20+\dfrac{1}{3}=\dfrac{61}{3}\)
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 \(x-\frac{4}{7}=0\Leftrightarrow x=\frac{4}{7}\)
Câu 2 :
\(x-y=7\)
\(\Rightarrow x=7+y\)
*)
\(B=\dfrac{3\left(7+y\right)-7}{2\left(7+y\right)+y}-\dfrac{3y+7}{2y+7+y}\)
\(=\dfrac{21+3y-7}{14+3y}-\dfrac{3y+7}{3y+7}\)
\(=\dfrac{14y+3y}{14y+3y}-1\)
\(=1-1\)
\(=0\)
Vậy B = 0
2/ Ta có :
\(B=\dfrac{3x-7}{2x+y}-\dfrac{3y+7}{2y+x}\)
\(=\dfrac{3x-\left(x-y\right)}{2x+y}-\dfrac{3y+\left(x-y\right)}{2y+x}\)
\(=\dfrac{3x-x+y}{2y+x}-\dfrac{3y+x-y}{2y+x}\)
\(=\dfrac{2x+y}{2x+y}-\dfrac{2y+x}{2y+x}\)
\(=1-1=0\)
\(\text{Câu 1 :}\)
\(A=\dfrac{5}{17}+\dfrac{-4}{9}-\dfrac{20}{31}+\dfrac{12}{17}-\dfrac{11}{31}\\ A=\left(\dfrac{5}{17}+\dfrac{12}{17}\right)-\left(\dfrac{20}{31}+\dfrac{11}{31}\right)+\dfrac{-4}{9}\\ A=1-1+-\dfrac{4}{9}\\ A=-\dfrac{4}{9}\)
\(B=\dfrac{-3}{7}+\dfrac{7}{15}+\dfrac{-4}{7}+\dfrac{8}{15}-\dfrac{-2}{3}\\ B=\left(\dfrac{-3}{7}+\dfrac{-4}{7}\right)+\left(\dfrac{7}{15}+\dfrac{8}{18}\right)-\dfrac{-2}{3}\\ B=\left(-1\right)+1+\dfrac{2}{3}\\ B=\dfrac{2}{3}\)
\(\text{Câu 2 : }\)
\(A< \dfrac{x}{9}\le B\\ \Rightarrow\dfrac{-4}{9}< \dfrac{x}{9}\le\dfrac{2}{3}\\ \Rightarrow\dfrac{-4}{9}< \dfrac{x}{9}\le\dfrac{6}{9}\\ \Rightarrow-4< x\le6\\ \Rightarrow x\in\left\{\pm4;\pm3;\pm2;\pm1;0;5;6\right\}\)
Mk nhầm chút nhé..
x không bằng -4 nhé. Nếu x bằng -4 thì bài sẽ như thế này:
\(-4\le x\le6\)