tìm x bt
b,(2+x)÷5=6
c,2.x—3=11
d,2+x÷5=6
a (2+x):5=6
b 2+x:5=6
c 2x-3=11
d 3.(x-18)+75=0
a) (2 + x) : 5 = 6
2 + x = 6 $\times$ 5
2 + x = 30
x = 30 - 2
x = 28
b) 2 + x : 5 = 6
x : 5 = 6 - 2
x : 5 = 4
x = 4 $\times$ 5
x = 20
c) 2x - 3 = 11
2x = 11 + 3
2x = 14
x = 14 : 2
x = 7
d) 3 . (x - 18) + 75 = 0
3 . (x - 18) = -75
x - 18 = -75 : 3
x - 18 = -25
x = -25 + 18
x = -7
Bài 3:
a) \(\dfrac{2x-1}{5}\)-\(\dfrac{x-2}{3}\)
=\(\dfrac{x+7}{15}\)
b) \(\dfrac{x+3}{2}\)-\(\dfrac{x-1}{3}\)
=\(\dfrac{x+5}{6}\)+1
c) \(\dfrac{2\left(x+5\right)}{3}\)+\(\dfrac{x+12}{2}\)
-\(\dfrac{5\left(x-2\right)}{6}\)=\(\dfrac{x}{3}\)+11
d) \(\dfrac{x-4}{5}\)+\(\dfrac{3x-2}{10}\)-x
=\(\dfrac{2x-5}{3}\)-\(\dfrac{7x+2}{6}\)
e) \(\dfrac{\left(2x-3\right)\left(2x+3\right)}{8}\)
=\(\dfrac{\left(x-4^{ }\right)^2}{6}\)+\(\dfrac{\left(x-2\right)^2}{3}\)
d) \(\dfrac{7x^2-14x-5}{15}\)
=\(\dfrac{\left(2x+1\right)^2}{5}\)-\(\dfrac{\left(x-1\right)^2}{3}\)
e) \(\dfrac{\left(7x+1\right)\left(x-2\right)}{10}\)+\(\dfrac{2}{5}\)
=\(\dfrac{\left(x-2\right)^2}{5}\)+\(\dfrac{\left(x-1\right)\left(x-3\right)}{2}\)
a) Ta có: \(\dfrac{2x-1}{5}-\dfrac{x-2}{3}=\dfrac{x+7}{15}\)
\(\Leftrightarrow\dfrac{3\left(2x-1\right)}{15}-\dfrac{5\left(x-2\right)}{15}=\dfrac{x+7}{15}\)
\(\Leftrightarrow6x-3-5x+10-x-7=0\)
\(\Leftrightarrow0x=0\)(luôn đúng)
Vậy: S={x|\(x\in R\)}
4) tìm số tự nhiên bt rằng
a) 4^11*25^11< 2^n*5^n < 20^12 * 5^12
b)4^5+4^5+4^5+4^5/3^5+3^5+3^5* 6^5+6^5+6^5+6^5+6^5+6^5/2^5+2^2=2n
5) Tìm x bt: x*(6-x)2003=(6-x)2003
6) Tìm x và y bt: (3x-5)^100+(2y+1)^200< 0
7)Tìm hai số tự nhiên m,n bt: 2m+2n=2m+n
8) Tìm các số nguyên x và y sao chp: (x+2)2+2(y-3)2<3
nhanh cái nào hehe
Tìm x, biết:
a, (x+8).(x+6)-x^2=104
b, (x+1).(x+2)-(x-3).(x+4)=6
c, 3.(2x-1).(x+2)-2.(3x+2).(x-4)=5
a: \(\Leftrightarrow14x=56\)
hay x=4
B1:chứng minh bt luôn dương
a/ x^4 + x^2 + 2
b/ (x+3)(x-11) + 2003
B2:cm bt luôn âm
a/-9x^2 + 12x - 15
b/-5-(x-1)(x+2)
B3:tìm gt nhỏ nhất của các bt sau:
a/A=11-10x-x^2
b/B=(x-2)(x-5)(x^2-7x-10)
B4:tìm x
a/(x+3)^3-x(3x+1)^2+(2x+1)(4x^2-2x+1)-3x^2=54
b/(x+3)^3-(x-3)(x^2+3x+9)+6(x+1)^2+3x^2=-33
a/x^4 lớn hơn hoặc = 0
x^2 lớn hơn hoặc = 0
2 > 0
=> x^4+x^2+2 >0 => bieu thức luôn dương
b/ (x+3)(x-11)+2003 <=> x^2 -8x -33 +2003 <=> x^2 -8x +1970 <=> x^2-8x+16+1954 <=> (x-4)^2+1954
ta có : (x-4)^2 lớn hơn hoặc = 0
1954 >0
=> (x-4)^2+1954>0 => bt luôn dương
Bài 1 trước nha . chúc bạn học tốt . Ủng hộ nha
\(=>-9\left(x^2-\frac{4}{3}x+\frac{5}{3}\right)=>-9\left(x^2-2.\frac{2}{3}x+\frac{4}{9}+\frac{11}{9}\right)=>-9\left(x-\frac{2}{3}\right)^2-11\)
Ta có \(\left(x-\frac{2}{3}\right)^2\ge0=>-9\left(x-\frac{2}{3}\right)^2\le0,-11< 0\)
\(-9\left(x-\frac{2}{3}\right)^2-11\le0\)=> bt luôn âm
\(=>-5-x^2-x+2=>-x^2-x-3=>-x^2-2.\frac{1}{2}x+\frac{1}{4}-\frac{13}{4}\)\(=>-\left(x-\frac{1}{2}\right)^2-\frac{13}{4}\)
Ta có \(\left(x-\frac{1}{2}\right)^2\ge0=>=>-\left(x-\frac{1}{2}\right)^2\le0,-\frac{13}{4}< 0\)
\(=>-\left(x-\frac{1}{2}\right)^2-\frac{13}{4}< 0\)=> bt luôn âm
ùng hộ mình nha. cảm ơn
Tìm x bt:
-1/2.x+2/3:(-4/5)=3/5-5/6
-1/2 .x + 2/3 : (-4/5) = 3/5 - 5/6
-1/2.x - 5/6 = -7/30
-1/2.x = -7/30 + 5/6
-1/2.x = 3/5
x = 3/5 : (-1/2)
x = 6/5
a)|-x+2/5|+1/2=3,5 b)21/5+3:|x/4-2/3|=6
c)7,5-3|5-2x|=-4,5 d)1/3-|5/4-2x|=1/4
e)21/5+3:|x/4-2/3|=6
a)|-x+2/5|+1/2=3,5 b)21/5+3:|x/4-2/3|=6
c)7,5-3|5-2x|=-4,5 d)1/3-|5/4-2x|=1/4
e)21/5+3:|x/4-2/3|=6
a: Ta có: \(\left|\dfrac{2}{5}-x\right|+\dfrac{1}{2}=3.5\)
\(\Leftrightarrow\left|x-\dfrac{2}{5}\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{2}{5}=3\\x-\dfrac{2}{5}=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{17}{5}\\x=-\dfrac{13}{5}\end{matrix}\right.\)
b: Ta có: \(\dfrac{21}{5}+3:\left|\dfrac{x}{4}-\dfrac{2}{3}\right|=6\)
\(\Leftrightarrow3:\left|\dfrac{1}{4}x-\dfrac{2}{3}\right|=6-\dfrac{21}{5}=\dfrac{9}{5}\)
\(\Leftrightarrow\left|\dfrac{1}{4}x-\dfrac{2}{3}\right|=\dfrac{5}{3}\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{4}x-\dfrac{2}{3}=\dfrac{5}{3}\\\dfrac{1}{4}x-\dfrac{2}{3}=-\dfrac{5}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{4}x=\dfrac{7}{3}\\\dfrac{1}{4}x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{28}{3}\\x=-4\end{matrix}\right.\)
tìm x bt
a) \(\dfrac{2}{3}\)x - \(\dfrac{7}{6}\) = \(\dfrac{12}{7}\) - \(\dfrac{1}{2}\)
b) ( \(1\dfrac{1}{2}\) + \(\dfrac{5}{3}\) - \(\dfrac{1}{6}\) ) : x = \(\dfrac{3}{4}\) - \(\dfrac{1}{2}\)
Lời giải:
a.
$\frac{2}{3}x-\frac{7}{6}=\frac{12}{7}-\frac{1}{2}=\frac{17}{14}$
$\frac{2}{3}x=\frac{17}{14}+\frac{7}{6}=\frac{50}{21}$
$x=\frac{50}{21}: \frac{2}{3}=\frac{25}{7}$
b.
$(1\frac{1}{2}+\frac{5}{3}-\frac{1}{6}):x=\frac{3}{4}-\frac{1}{2}$
$3:x=\frac{1}{4}$
$x=3: \frac{1}{4}=12$