/6,5-x/ =35
1\3+2\3 :x=-1+/-2\3/
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Theo đề bài, ta có:
x:y:z=2/3:3/5:1/2
=> x/2/3=y/3/5=z/1/2
và x-y=-6,5
Theo tính chất dãy tỉ số bằng nhau, ta có:
x/2/3=y/3/5=x-y/2/3-3/5=-6,5/1/15=-13/30
=> x/2/3=-13/30=>x=-13/30.2/3=-13/45
=>y/3/5=-13/45=>y=-13/45.3/5=-13/75
=>z/1/2=-13/45=>z=-13/45.1/2=-13/90
Vậy x=-13/45 ; y=-13/75 ; z=-13/90
Tick cho mình nha!
x:y:z=2/3:3/5:1/2
=>\(\dfrac{x}{\dfrac{2}{3}}=\dfrac{y}{\dfrac{3}{5}}=\dfrac{z}{\dfrac{1}{2}}\)=\(\dfrac{x-y}{\dfrac{2}{3}-\dfrac{3}{5}}=\dfrac{6,5}{\dfrac{1}{15}}=97,5\)
=>\(\dfrac{x}{\dfrac{2}{3}}=97,5\Rightarrow x=65;\dfrac{y}{\dfrac{3}{5}}=97,5\Rightarrow y=58,5\)
\(\dfrac{z}{\dfrac{1}{2}}=97,5\Rightarrow z=48,75\)
vậy .......
1/\(\left|3x+2\right|+\left|9x^2-4\right|=0\)
<=> \(\hept{\begin{cases}\left|3x+2\right|=0\\\left|9x^2-4\right|=0\end{cases}}\)
<=> \(\hept{\begin{cases}3x+2=0\\9x^2-4=0\end{cases}}\)
<=> \(\hept{\begin{cases}x=-\frac{2}{3}\\x=\frac{2}{3}\end{cases}}\)
<=> \(x\in\varnothing\)
2/ \(\left|x-5\right|+\left|x-25\right|=0\)
<=> \(\hept{\begin{cases}\left|x-5\right|=0\\\left|x-25\right|=0\end{cases}}\)
<=> \(\hept{\begin{cases}x=5\\x=25\end{cases}}\)
<=> \(x\in\varnothing\)
3/ \(\left|2x\right|-\left|-3,5\right|=\left|-6,5\right|\)
<=> \(\left|2x\right|-3,5=6,5\)
<=> \(\left|2x\right|=10\)
<=> \(2x=\pm10\)
<=> \(x=\pm5\)
4/ \(\frac{5}{3}-\left|x-\frac{1}{3}\right|=\frac{1}{3}\)
<=> \(-\left|x-\frac{1}{3}\right|=-\frac{4}{3}\)
<=> \(\left|x-\frac{1}{3}\right|=\frac{4}{3}\)
<=> \(\orbr{\begin{cases}x-\frac{1}{3}=\frac{4}{3}\\x-\frac{1}{3}=-\frac{4}{3}\end{cases}}\)
<=> \(\orbr{\begin{cases}x=\frac{5}{3}\\x=-1\end{cases}}\)
l3x + 2l +l9x2 - 4l = 0
=> l3x + 2l =0 hoặc l9x2-4l =0
=> 3x + 2 = 0 9x2-4 =0
=> 3x = -2 9x2 =4
=> x = -2:3 x2 = 4:9
=> x = -2/3 x2 =4/9
=> x =2/3
Vậy x ={-2/3 ; 2/3}
câu 2 là tương tự
Bài 2:
a:
1: \(\dfrac{a}{b}=\dfrac{c}{d}\)
\(\Leftrightarrow\dfrac{a+b}{a}=\dfrac{c+d}{c}\)
hay \(\dfrac{a}{a+b}=\dfrac{c}{c+d}\)
a) Ta có: \(9\cdot5^x=6\cdot5^6+3\cdot5^6\)
\(\Leftrightarrow9\cdot5^x=9\cdot5^6\)
\(\Leftrightarrow5^x=5^6\)
hay x=6
b) Ta có: \(2^{2x+1}+4^{x+3}=264\)
\(\Leftrightarrow4^x\cdot2+4^x\cdot64=264\)
\(\Leftrightarrow4^x=4\)
hay x=1
Bài 3:
b: \(x^2+2x+1=\left(x+1\right)^2\)
c: \(x^2-16=\left(x-4\right)\left(x+4\right)\)
d: \(\left(2x-1\right)^2-\left(x+3\right)^2\)
\(=\left(2x-1-x-3\right)\left(2x-1+x+3\right)\)
\(=\left(x-4\right)\left(3x+2\right)\)
a) $|6,5 - x| = 35$
$6,5 - x= 35$ hoặc $6,5 - x = -35$
$x = -28,5$ hoặc $x = 41,5$.
b) $\dfrac13 + \dfrac23$ : $x = -1 + \left|-\dfrac23\right|$
$\dfrac13 + \dfrac23$ : $x = -1 + \dfrac23$
$\dfrac23$ : $x = -\dfrac23$
$x = -1$.