tìm x,biết
a) /x/=1/5
tìm x , biết
a) 17/6- x( x-7/6)= 7/4
b) 3/35 - ( 3/5-x)= 2/7
tìm x thuộc Z , biết
3/4-5/6 < x/12 < 1 -( 2/3-1/4)
tìm x biết
a ) 2x-3=x + 1/2
b) 4x- ( x+ 1/2) = 2x - ( 1/2 - 5 )
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 \(x=\dfrac{3}{5}-\dfrac{-1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)
Bài 3:
a) Ta có: \(2x-3=x+\dfrac{1}{2}\)
\(\Leftrightarrow2x-x=\dfrac{1}{2}+3\)
\(\Leftrightarrow x=\dfrac{7}{2}\)
b) Ta có: \(4x-\left(x+\dfrac{1}{2}\right)=2x-\left(\dfrac{1}{2}-5\right)\)
\(\Leftrightarrow3x-\dfrac{1}{2}-2x+\dfrac{1}{2}-5=0\)
\(\Leftrightarrow x=5\)
Tìm x, biết
a. (x+1)(x+3)-(x-2)(x+5) = 2
Mình cần gấp!!!!!!!!!!!
\(\left(x+1\right)\left(x+3\right)-\left(x-2\right)\left(x+5\right)=2\)
\(\Leftrightarrow x^2+4x+3-x^2-3x+10=2\)
\(\Leftrightarrow x=-11\)
tìm x biết
a) 3x(x-1)+5(x-1)=0
b) x2-3x+2=0
a) \(\Rightarrow\left(x-1\right)\left(3x+5\right)=0\Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{5}{3}\end{matrix}\right.\)
b) \(\Rightarrow x\left(x-1\right)-2\left(x-1\right)=0\Rightarrow\left(x-1\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
tìm x biết
a, -5 (x^2 - 3x + 1 ) + x ( 1 +5x) = x-2
b,-4x ( x - 5 ) + 7x (x - 4 ) - 3x^ 2 = 12
Lời giải:
a.
PT $\Leftrightarrow -5x^2+15x-5+x+5x^2=x-2$
$\Leftrightarrow 16x-5=x-2$
$\Leftrightarrow 15x=3$
$\Leftrightarrow x=\frac{3}{15}=\frac{1}{5}$
b.
PT $\Leftrightarrow -4x^2+20x+7x^2-28x-3x^2=12$
$\Leftrightarrow -8x=12$
$\Leftrightarrow x=\frac{-3}{2}$
1.Tìm x,biết
a.5/7+4/3:x=1/7
b.5/3xX-1/4=2/6
a. \(\dfrac{5}{7}+\dfrac{4}{3}:x=\dfrac{1}{7}\)
<=> \(\dfrac{5}{7}+\dfrac{4}{3}.\dfrac{1}{x}=\dfrac{1}{7}\)
<=> \(\dfrac{5}{7}+\dfrac{4}{3x}=\dfrac{1}{7}\) ĐKXĐ: x \(\ne\) 0
<=> \(\dfrac{15x}{21x}+\dfrac{28}{21x}=\dfrac{3x}{21x}\)
<=> 15x + 28 = 3x
<=> 15x - 3x = -28
<=> 12x = -28
<=> x = \(\dfrac{-28}{12}=-\dfrac{7}{3}\)
b. \(\dfrac{5}{3}x.\dfrac{-1}{4}=\dfrac{2}{6}\)
<=> \(\dfrac{-5x}{12}=\dfrac{2}{6}\)
<=> -5x . 6 = 12 . 2
<=> -30x = 24
<=> x = \(-\dfrac{4}{5}\)
Bài 1: Tìm số nguyên x biết
a) x/4= 1/x
b) 1/5= x:4-1/10
a) \(\dfrac{x}{4}\)=\(\dfrac{1}{x}\)→x2=4→x=2 hoặc x=-2
b) \(\dfrac{1}{5}\) = x:4-\(\dfrac{1}{10}\)→x:4=\(\dfrac{1}{5}\)+\(\dfrac{1}{10}\)→x:4=\(\dfrac{2+1}{10}\)→x:4=\(\dfrac{3}{10}\)→x=\(\dfrac{3}{10}\)x4→x=\(\dfrac{12}{10}\)→x=\(\dfrac{6}{5}\)
1. Tìm số nguyên x, biết
a) 3^−1.4^x+3.4^x=5/3.2^7
tìm số nguyên x biết
a, x/2=8/x
b,x+1/5=x+1/5
c,x+1/5=x+3/10
d,x/4=18/x+1
\(a,\dfrac{x}{2}=\dfrac{8}{x}\\ \Rightarrow x^2=16\\ \Rightarrow x=\pm4\\ b,\dfrac{x+1}{5}=\dfrac{x+1}{5}\left(luôn.đúng\right)\\ c,\dfrac{x+1}{5}=\dfrac{x+3}{10}\\ \Rightarrow\dfrac{2x+2}{10}=\dfrac{x+3}{10}\\ \Rightarrow2x+2=x+3\\ \Rightarrow2x-x=3-2\\ \Rightarrow x=1\\ d,\dfrac{x}{4}=\dfrac{18}{x+1}\\ \Rightarrow x\left(x+1\right)=4.18\\ \Rightarrow x^2+x=72\\ \Rightarrow x^2+x-72=0\\ \Rightarrow\left(x^2+9x\right)-\left(8x+72\right)=0\\ \Rightarrow x\left(x+9\right)-8\left(x+9\right)=0\\ \Rightarrow\left(x-8\right)\left(x+9\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=8\\x=-9\end{matrix}\right.\)
bài 1: Tìm x biết
a) 2.x+(-5)=-18
b) 3.3=81
c) 64-4.(5-x)=40
a: =>2x=-18+5=-13
=>x=-13/2
b: =>3^x-1=81
=>x-1=4
=>x=5
c: =>4(5-x)=24
=>5-x=6
=>x=-1
Bài 1 tìm số nguyên x,y biết
a, x/3=y/7
b,x/y-1=5/-19
\(\dfrac{x}{3}=\dfrac{y}{7}\Rightarrow\dfrac{x}{y}=\dfrac{3}{7}\)
\(\dfrac{x}{y}-1=\dfrac{-5}{19}\Rightarrow\dfrac{x}{y}=\dfrac{14}{19}\)
Vô lí => không có x,y thỏa mãn
a) Ta có: \(\dfrac{x}{3}=\dfrac{y}{7}\)
nên \(\dfrac{x}{y}=\dfrac{3}{7}\)
b) Ta có: \(\dfrac{x}{y-1}=\dfrac{5}{-19}\)
\(\Leftrightarrow\dfrac{x}{5}=\dfrac{y-1}{-19}\)
hay \(\dfrac{x}{5}=\dfrac{1-y}{19}\)