3^x-1+3^x+1=810
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Đặt x/2=y/3=z/5=k
=>x=2k; y=3k; z=5k
xyz=810
\(\Leftrightarrow30k^3=810\)
=>k=3
=>x=6; y=9; z=15
Lời giải:
$x-\frac{x}{3}\times \frac{3}{2}=2-\frac{1}{2}$
$x-x\times \frac{1}{2}=\frac{3}{2}$
$x\times (1-\frac{1}{2})=\frac{3}{2}$
$x\times \frac{1}{2}=\frac{3}{2}$
$x=\frac{3}{2}: \frac{1}{2}=3$
\(\dfrac{1+x}{1-x}+3=\dfrac{x-3}{x-1}\)
\(ĐK:x\ne1\)
\(\Leftrightarrow\dfrac{1+x}{1-x}+3=\dfrac{3-x}{1-x}\)
\(\Leftrightarrow\dfrac{\left(1+x\right)+3\left(1-x\right)}{1-x}=\dfrac{3-x}{1-x}\)
\(\Leftrightarrow\left(1+x\right)+3\left(1-x\right)=3-x\)
\(\Leftrightarrow1+x+3-3x=3-x\)
\(\Leftrightarrow-x=-1\)
\(\Leftrightarrow x=1\left(ktm\right)\)
Vậy pt vô nghiệm
\(\dfrac{1+x}{1-x}+3=\dfrac{x-3}{x-1}\) đề như thế này phải ko?
\(\Leftrightarrow2\left(x+1\right)^3=56\Leftrightarrow\left(x+1\right)^3=28\Leftrightarrow\)
a) \(\dfrac{13}{20}+\dfrac{3}{5}+x=\dfrac{5}{6}\)
\(\Rightarrow\dfrac{5}{4}+x=\dfrac{5}{6}\)
\(\Rightarrow x=\dfrac{5}{6}-\dfrac{5}{4}\)
\(\Rightarrow x=\dfrac{-5}{12}\)
b) \(x+\dfrac{1}{3}=\dfrac{2}{5}-\dfrac{-1}{3}\)
\(\Rightarrow x+\dfrac{1}{3}=\dfrac{11}{15}\)
\(\Rightarrow x=\dfrac{11}{15}-\dfrac{1}{3}\)
\(\Rightarrow x=\dfrac{2}{5}\)
c)\(\dfrac{-5}{8}-x=\dfrac{-3}{20}-\dfrac{-1}{6}\)
\(\dfrac{-5}{8}-x=\dfrac{1}{60}\)
\(\Rightarrow x=\dfrac{-5}{8}-\dfrac{1}{60}\)
\(\Rightarrow x=\dfrac{-77}{120}\)
d) \(\dfrac{3}{5}-x=\dfrac{1}{4}+\dfrac{7}{10}\)
\(\Rightarrow\dfrac{3}{5}-x=\dfrac{19}{20}\)
\(\Rightarrow x=\dfrac{3}{5}-\dfrac{19}{20}\)
\(\Rightarrow x=\dfrac{-7}{20}\)
e) \(\dfrac{-3}{7}-x=\dfrac{4}{5}+\dfrac{-2}{3}\)
\(\Rightarrow\dfrac{-3}{7}-x=\dfrac{2}{15}\)
\(\Rightarrow x=\dfrac{-3}{7}-\dfrac{2}{15}\)
\(\Rightarrow x=\dfrac{-59}{105}\)
g) \(\dfrac{-5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)
\(\Rightarrow\dfrac{-5}{6}-x=\dfrac{1}{4}\)
\(\Rightarrow x=\dfrac{-5}{6}-\dfrac{1}{4}\)
\(\Rightarrow x=\dfrac{-13}{12}\)
a) x - 1/2 = 3/5
x = 3/5 + 1/2
x = 11/10
b) x - 1/2 = -2/3
x = -2/3 + 1/2
x = -1/6
c) 2/5 - x = 0,25
x = 2/5 - 0,25
x = 2/5 - 1/4
x = 3/20
\(2x\left(x+3\right)-3\left(x^2+1\right)=x+1-x\left(x-2\right)\)
\(\Leftrightarrow2x^2+6x-3x^2-3=x+1-x^2+2x\)
\(\Leftrightarrow-x^2+6x-3=-x^2+3x+1\)
\(\Leftrightarrow3x=4\)
hay \(x=\dfrac{4}{3}\)
\(2x\left(x+3\right)-3\left(x^2+1\right)=x+1-x\left(x-2\right)\)
\(\Leftrightarrow2x^2+6x-3x^2-3=x+1-x^2+2x\)
\(\Leftrightarrow3x=4\Leftrightarrow x=\dfrac{4}{3}\)
Đề trước đó:
(x-7)(x+1)-(x-3)^2=(3x-5)(3x+5)-(3x+1)^2+(x-2)^2-x
<=>x^2+x-7x-7-x^2+6x-9=9x^2-25-9x^2-6x-1+x^2-4x+4-x
<=>x^2-11x-6=0
<=>x^2-2x. 11/2 + 121/4-145/4=0
<=>(x-11/2)^2=145/4
<=>|x-11/2|=căn(145)/2
<=>x=[11+-căn(145)]/2
x=5 bạn nhé
Bạn làm ngắn gọn quá nên mk chưa hiểu rõ ạ! Bạn có thể trình bày ra đc không ạ?