Bài 1 Tìm số nguyên a để
\(\dfrac{2a+9}{a+3}\) + \(\dfrac{5a+17}{a+3}\)- \(\dfrac{3a}{a+3}\)là số nguyên
Bài 2 Tìm x
\(\dfrac{x+2}{327}\)+\(\dfrac{x+3}{326}\)+\(\dfrac{x+4}{325}\)+\(\dfrac{x+5}{324}\)+\(\dfrac{x+349}{5}\)=0
a)Tìm x:\(\dfrac{x+2}{327}+\dfrac{x+3}{326}+\dfrac{x+4}{325}+\dfrac{x+5}{324}+\dfrac{x+349}{5}=0\)
\(\dfrac{x+2}{327}+\dfrac{x+3}{326}+\dfrac{x+4}{325}+\dfrac{x+5}{324}+\dfrac{x+349}{5}=0\)
\(\Rightarrow\left(\dfrac{x+2}{327}+1\right)+\left(\dfrac{x+3}{326}+1\right)+\left(\dfrac{x+4}{325}+1\right)+\left(\dfrac{x+5}{324}+1\right)+\left(\dfrac{x+349}{5}-4\right)=0\)
\(\Rightarrow\dfrac{x+329}{327}+\dfrac{x+329}{326}+\dfrac{x+329}{325}+\dfrac{x+329}{324}+\dfrac{x+329}{5}=0\)
\(\Rightarrow\left(x+329\right)\left(\dfrac{1}{327}+\dfrac{1}{326}+\dfrac{1}{325}+\dfrac{1}{324}+\dfrac{1}{5}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+329=0\\\dfrac{1}{327}+\dfrac{1}{326}+\dfrac{1}{325}+\dfrac{1}{324}+\dfrac{1}{5}=0\left(vôlí\right)\end{matrix}\right.\)
\(\Rightarrow x=-329\)
\(\dfrac{x+2}{327}+\dfrac{x+3}{326}+\dfrac{x+4}{325}+\dfrac{x+5}{324}+\dfrac{x+349}{5}=0\)
⇔ \(\left(\dfrac{x+2}{327}+1\right)+\left(\dfrac{x+3}{326}+1\right)+\left(\dfrac{x+4}{325}+1\right)+\left(\dfrac{x+5}{324}+1\right)+\)
\(\left(\dfrac{x+349}{5}-4\right)=0\)
⇔ \(\dfrac{x+329}{327}+\dfrac{x+329}{326}+\dfrac{x+329}{325}+\dfrac{x+329}{324}+\dfrac{x+329}{5}=0\)
⇔ \(\left(x+329\right)\left(\dfrac{1}{327}+\dfrac{1}{326}+\dfrac{1}{325}+\dfrac{1}{324}+\dfrac{1}{5}\right)=0\)
⇔ \(x+329=0\) Vì \(\left(\dfrac{1}{327}+\dfrac{1}{326}+\dfrac{1}{325}+\dfrac{1}{324}+\dfrac{1}{5}\right)\) ≠ 0
⇔ \(x=-329\)
Bài 9: Tìm \(x\) biết
1) \(\dfrac{9}{x}=\dfrac{-35}{105}\) 2) \(\dfrac{12}{5}=\dfrac{32}{x}\) 3) \(\dfrac{x}{2}=\dfrac{32}{x}\) 4) \(\dfrac{x-2}{4}=\dfrac{x-1}{5}\)
Bài 10: Cho biểu thức \(A=\dfrac{3}{n+2}\)
a) Số nguyên n phài thỏa mãn điều kiện gì để A là phân số? b) Tìm phân số A khi n = 0? n = 2? n = -7?
Bài 10:
a: Để A là phân số thì n+2<>0
hay n<>-2
b: Khi n=0 thì A=3/2
Khi n=2 thì A=3/(2+2)=3/4
Khi n=-7 thì A=3/(-7+2)=-3/5
Bài 9:
1)9/x = -35/105 2) 12/5 = 32/x 3)x/2 = 32/x x = 9. (-35)/105 x.12/5 = x.32/x 2x.x/2 = 2x.32/x
x = -3 x.12/5=32 xx = 2.32
x= 32:12/5 x^2 = 2.32
x = 40/3 x^2 = 64
x = 8
4) x-2/4 = x-1/5
5(x-2) = 4(x-1)
5x - 10 = 4x - 4
5x - 4x = 10 - 4
x = 6
Bài 10:Cho biểu thức
bài 1
a> Tính giá tị của biểu thức A=\(x^2-3x+1\) khi \(\left|x+\dfrac{1}{3}\right|=\dfrac{2}{3}\)
b> Tìm x biết: \(\dfrac{3-x}{20}=\dfrac{-5}{x-3}\)
Bài 2
a> Tìm các số x,y thỏa mãn: \(\dfrac{x-1}{3}=\dfrac{y+2}{5}=\dfrac{x+y+1}{x-2}\)
b> Cho x nguyên, tìm giá trị lớn nhất của biểu thức sau: A=\(\dfrac{2x+1}{x-3}\)
c> Tìm số có 2 chữ số \(\overline{ab}\) biết: \(\left(\overline{ab}\right)^2\)=\(\left(a+b\right)^3\)
\(\overline{ab}\)
Bài 1:
b) ĐKXĐ: \(x\ne3\)
Ta có: \(\dfrac{3-x}{20}=\dfrac{-5}{x-3}\)
\(\Leftrightarrow\dfrac{x-3}{-20}=\dfrac{-5}{x-3}\)
\(\Leftrightarrow\left(x-3\right)^2=100\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=10\\x-3=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=13\left(nhận\right)\\x=-7\left(nhận\right)\end{matrix}\right.\)
Vậy: \(x\in\left\{13;-7\right\}\)
tìm x e Q
a) \(\dfrac{x+1}{3}+\dfrac{x+1}{4}+\dfrac{x+1}{5}=\dfrac{x+1}{6}\)
b) \(\dfrac{x+1}{2020}+\dfrac{x+2}{2019}=\dfrac{x+3}{2018}+\dfrac{x+4}{2017}\)
c) \(\dfrac{x+2}{327}+\dfrac{x+3}{326}+\dfrac{x+4}{325}+\dfrac{x+5}{324}+\dfrac{x+349}{5}=0\)
\(\dfrac{x+1}{3}+\dfrac{x+1}{4}+\dfrac{x+1}{5}=\dfrac{x+1}{6}\)
\(\dfrac{x+1}{3}+\dfrac{x+1}{4}+\dfrac{x+1}{5}-\dfrac{x+1}{6}=0\)
\(\left(x+1\right)\left(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}\right)=0\)
\(\)vì \(\dfrac{1}{3}>\dfrac{1}{6};\dfrac{1}{4}>\dfrac{1}{6};\dfrac{1}{5}>\dfrac{1}{6}=>\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}>0\)
\(=>x+1=0\)
\(=>x=-1\)
b,
\(\dfrac{x+1}{2020}+\dfrac{x+2}{2019}=\dfrac{x+3}{2018}+\dfrac{x+4}{2017}\)
\(\left(\dfrac{x+1}{2020}+1\right)+\left(\dfrac{x+2}{2019}+1\right)=\left(\dfrac{x+3}{2018}+1\right)+\left(\dfrac{x+4}{2017}+1\right)\)
\(\dfrac{x+2021}{2020}+\dfrac{x+2021}{2019}=\dfrac{x+2021}{2018}+\dfrac{x+2021}{2017}\)
\(=>\dfrac{x+2021}{2020}+\dfrac{x+2021}{2019}-\dfrac{x+2021}{2018}-\dfrac{x+2021}{2017}=0\)
\(=>\left(x+2021\right)\left(\dfrac{1}{2020}+\dfrac{1}{2019}-\dfrac{1}{2018}-\dfrac{1}{2017}\right)=0\)
Vì \(\dfrac{1}{2020}< \dfrac{1}{2018};\dfrac{1}{2019}< \dfrac{1}{2017}=>\dfrac{1}{2020}+\dfrac{1}{2019}-\dfrac{1}{2018}-\dfrac{1}{2017}< 0\)
\(=>x+2021=0\)
\(=>x=-2021\)
c,
\(\dfrac{x+2}{327}+\dfrac{x+3}{326}+\dfrac{x+4}{325}+\dfrac{x+5}{324}+\dfrac{x+349}{5}=0\)
\(\left(\dfrac{x+2}{327}+1\right)+\left(\dfrac{x+3}{326}+1\right)+\left(\dfrac{x+4}{325}+1\right)+\left(\dfrac{x+5}{324}+1\right)+\left(\dfrac{x+349}{5}-4\right)=0\)
\(\dfrac{x+329}{327}+\dfrac{x+329}{326}+\dfrac{x+329}{325}+\dfrac{x+329}{324}+\dfrac{x+329}{5}=0\)
\(=>\left(x+329\right)\left(\dfrac{1}{327}+\dfrac{1}{326}+\dfrac{1}{325}+\dfrac{1}{324}+\dfrac{1}{5}\right)=0\)
Vì \(\dfrac{1}{327}+\dfrac{1}{326}+\dfrac{1}{325}+\dfrac{1}{324}+\dfrac{1}{5}>0\)
\(=>x+329=0\)
\(=>x=-329\)
bài 1: TBC của 2 số là 36, số bé là 17, tìm số lớn
bài 2: Tìm y
a) 4567 + y : 34 = 10987 b)\(\dfrac{4}{3}+\dfrac{1}{2}\): y = 2
bài 3 tính
\(\dfrac{2}{5}x\dfrac{2}{5}+\dfrac{9}{8}:3\)=....................................................................
2 - (\(\dfrac{1}{7}\) x 4 + \(\dfrac{5}{21}\)) =....................................................................
Bài 1:
Tổng của 2 số là
\(36\times2=72\)
Số lớn là
\(72-17=55\)
Bài 2:
a) \(4567+y\div34=10987\)
\(y\div34=10987-4567\)
\(y\div34=6420\)
\(y=6420\times34\)
\(y=218280\)
b) \(\dfrac{4}{3}+\dfrac{1}{2}\div y=2\)
\(\dfrac{1}{2}\div y=2-\dfrac{4}{3}\)
\(\dfrac{1}{2}\div y=\dfrac{2}{3}\)
\(y=\dfrac{1}{2}\div\dfrac{2}{3}\)
\(y=\dfrac{3}{4}\)
Bài 3:
a) \(\dfrac{2}{5}\times\dfrac{2}{5}+\dfrac{9}{8}\div3=\dfrac{4}{25}+\dfrac{9}{8}\times\dfrac{1}{3}=\dfrac{4}{25}+\dfrac{3}{8}=\dfrac{107}{200}\)
b) \(2-\left(\dfrac{1}{7}\times4+\dfrac{5}{21}\right)=2-\left(\dfrac{4}{7}+\dfrac{5}{21}\right)=2-\dfrac{17}{21}=\dfrac{25}{21}\)
Bài 1 : Gọi a là số lớn, b là số bé, theo đề bài ta có :
(a+b):2=36⇒a+b=72
mà b=17
Nên a=72-17=55
Bài 2 :
a) 4567+y:34=10987
⇒ y:34=10987-4567
⇒ y:34=6420
⇒ y=6420x34
⇒ y=218280
b) \(\dfrac{4}{3}+\dfrac{1}{2}:y=2\)
\(\Rightarrow\dfrac{1}{2}:y=2-\dfrac{4}{3}\)
\(\Rightarrow\dfrac{1}{2}:y=\dfrac{2}{3}\)
\(\Rightarrow y=\dfrac{1}{2}:\dfrac{2}{3}\)
\(\Rightarrow y=\dfrac{1}{2}x\dfrac{3}{2}\)
\(\Rightarrow y=\dfrac{3}{4}\)
Bài 3 :
\(\dfrac{2}{5}x\dfrac{2}{5}+\dfrac{9}{8}:3=\dfrac{4}{25}+\dfrac{9}{8}x\dfrac{1}{3}=\dfrac{4}{25}+\dfrac{3}{8}\)
= \(\dfrac{4x8}{25x8}+\dfrac{25x3}{25x8}=\dfrac{32}{200}+\dfrac{75}{200}=\dfrac{107}{200}\)
\(2-\left(\dfrac{1}{7}x4+\dfrac{5}{21}\right)=2-\left(\dfrac{4}{7}+\dfrac{5}{21}\right)=2-\left(\dfrac{12}{21}+\dfrac{5}{21}\right)=2-\dfrac{17}{21}=\dfrac{42}{21}-\dfrac{17}{21}=\dfrac{25}{21}\)
Bài 4: tìm x:
a) \(\dfrac{4}{3}\) + (1,25 - x) = 2,25
b) \(\dfrac{17}{6}\) - (x - \(\dfrac{7}{6}\) ) = \(\dfrac{7}{4}\)
c) 4 - (2x + 1) = 3 - \(\dfrac{1}{3}\)
bài 15:
a) (\(\dfrac{-2}{3}\))9 : x = (\(\dfrac{-2}{3}\))
b) x : (\(\dfrac{4}{9}\))5 = (\(\dfrac{4}{9}\))4
c) (x + 4)3 = -125
d) (10 - 5x)3 = 64
e) (4x + 5)2 = 81
Bài 16:
a) 4 - \(1\dfrac{2}{5}\) - \(\dfrac{8}{3}\)
b) -0,6 - \(\dfrac{-4}{9}\) - \(\dfrac{16}{15}\)
c) \(-\dfrac{15}{4}\) . (\(\dfrac{-7}{15}\)) . (\(-2\dfrac{2}{5}\)
Gi ải gấp giúp mình ạ, mình rất cần gấp
Bài 4:
a) \(\dfrac{4}{3}+\left(1,25-x\right)=2,25\)
\(1,25-x=2,25-\dfrac{4}{3}=\dfrac{9}{4}-\dfrac{4}{3}\)
\(1,25-x=\dfrac{11}{12}\)
\(x=1,25-\dfrac{11}{12}=\dfrac{5}{4}-\dfrac{11}{12}\)
\(x=\dfrac{1}{3}\)
b) \(\dfrac{17}{6}-\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)
\(x-\dfrac{7}{6}=\dfrac{17}{6}-\dfrac{7}{4}=\dfrac{34}{12}-\dfrac{21}{12}\)
\(x-\dfrac{7}{6}=\dfrac{13}{12}\)
\(x=\dfrac{13}{12}+\dfrac{7}{6}=\dfrac{13}{12}+\dfrac{14}{12}\)
\(x=\dfrac{27}{12}=\dfrac{9}{4}\)
c) \(4-\left(2x+1\right)=3-\dfrac{1}{3}=\dfrac{9}{3}-\dfrac{1}{3}\)
\(4-\left(2x+1\right)=\dfrac{8}{3}\)
\(2x+1=\dfrac{8}{3}+4=\dfrac{8}{3}+\dfrac{12}{3}\)
\(2x+1=\dfrac{20}{3}\)
\(2x=\dfrac{20}{3}-1=\dfrac{20}{3}-\dfrac{3}{3}\)
\(2x=\dfrac{17}{3}\)
\(x=\dfrac{17}{3}.\dfrac{1}{2}=\dfrac{17}{6}\)
Bài 15:
a) \(\left(\dfrac{-2}{3}\right)^9:x=\dfrac{-2}{3}\)
\(x=\left(\dfrac{-2}{3}\right)^9:\dfrac{-2}{3}=\left(\dfrac{-2}{3}\right)^{9-1}\)
\(=>x=\left(\dfrac{-2}{3}\right)^8\)
b) \(x:\left(\dfrac{4}{9}\right)^5=\left(\dfrac{4}{9}\right)^4\)
\(x=\left(\dfrac{4}{9}\right)^4.\left(\dfrac{4}{9}\right)^5=\left(\dfrac{4}{9}\right)^{4+5}\)
\(=>x=\left(\dfrac{4}{9}\right)^9\)
c) \(\left(x+4\right)^3=-125\)
\(\left(x+4\right)^3=\left(-5\right)^3\)
\(=>x+4=-5\)
\(x=-5-4\)
\(=>x=-9\)
d) \(\left(10-5x\right)^3=64\)
\(\left(10-5x\right)^3=4^3\)
\(=>10-5x=4\)
\(5x=10-4\)
\(5x=6\)
\(=>x=\dfrac{6}{5}\)
e) \(\left(4x+5\right)^2=81\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(4x+5\right)^2=\left(-9\right)^2\\\left(4x+5\right)^2=9^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+5=-9\\4x+5=9\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=-14\\4x=4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-14}{4}\\x=1\end{matrix}\right.\)
Bài 16:
a) \(4-1\dfrac{2}{5}-\dfrac{8}{3}\)
\(=4-\dfrac{7}{5}-\dfrac{8}{3}\)
\(=\dfrac{60-21-40}{15}=\dfrac{-1}{15}\)
b) \(-0,6-\dfrac{-4}{9}-\dfrac{16}{15}\)
\(=\dfrac{-3}{5}+\dfrac{4}{9}-\dfrac{16}{15}\)
\(=\dfrac{\left(-27\right)+20-48}{45}=\dfrac{-55}{45}=\dfrac{-11}{9}\)
c) \(-\dfrac{15}{4}.\left(\dfrac{-7}{15}\right).\left(-2\dfrac{2}{5}\right)\)
\(=\dfrac{7}{4}.\dfrac{-12}{5}\)
\(=\dfrac{-21}{5}\)
\(#Wendy.Dang\)
cho B= 1\ 2(n-1)^2 +3 để tìm GTLN
a, x+2\327 + x+3\326 + x+4\325 + x+5\324 + x+349\5 =0
b, 2a+9\a+3 + 5a+17\a+3 - 3a\a+3 là số nguyên
a.
Ta có:
(x+2)/327+(x+3)/326+(x+4)/325+(x+5)/324+(x+349)/5=0
<=>(x+2)/327+(x+3)/326+(x+4)/325+(x+5)/324+(x+329)-4 (giải thích: (x+349)/5=(x+329+20)/5=(x+329)/5+4)
<=>1+(x+2)/327+1+(x+3)/326+1+(x+4)/325+1+(x+5)324+(x+329)/5=0
<=>(x+329)/327+(x+329)/326+(x+329)/325+(x+329)/324+(x+329)/5=0
<=>x+329(1/327+1/326+1/325+1/324+1/5)=0
Vì (1/327+...+1/5) khác 0 => x+329=0
=>x=-329
Cho các biểu thức:\(A=\dfrac{2x}{x+3}+\dfrac{x+1}{x-3}+\dfrac{3-11x}{9-x^2};B=\dfrac{x-3}{x+1}\) \(\left(0\le x,x\ne9\right)\) a, Rút gọn A
b, Với P = A.B ,tìm x để P = \(\dfrac{9}{2}\)
c, Tìm x để B < 1
d, Tìm số nguyên x để P là số nguyên
a) Ta có: \(A=\dfrac{2x}{x+3}+\dfrac{x+1}{x-3}+\dfrac{3-11x}{9-x^2}\)
\(=\dfrac{2x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\dfrac{\left(x+1\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{11x-3}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{2x^2-6x+x^2+4x+3+11x-3}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{3x^2+9x}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{3x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3x}{x-3}\)
b)
ĐKXĐ: \(x\notin\left\{3;-3;-1\right\}\)
Ta có: P=AB
\(=\dfrac{3x}{x-3}\cdot\dfrac{x-3}{x+1}\)
\(=\dfrac{3x}{x+1}\)
Để \(P=\dfrac{9}{2}\) thì \(\dfrac{3x}{x+1}=\dfrac{9}{2}\)
\(\Leftrightarrow9\left(x+1\right)=6x\)
\(\Leftrightarrow9x-6x=-9\)
\(\Leftrightarrow3x=-9\)
hay x=-3(loại)
Vậy: Không có giá trị nào của x để \(P=\dfrac{9}{2}\)
Tìm x biết
a)\(\dfrac{x+2}{327}+\dfrac{x+3}{326}+\dfrac{x+4}{325}+\dfrac{x+5}{324}+\dfrac{x+349}{5}=0\)
\(\left(\dfrac{x+2}{327}+1\right)+\left(\dfrac{x+3}{326}+1\right)+\left(\dfrac{x+4}{325}+1\right)+\left(\dfrac{x+5}{324}\right)+\left(\dfrac{x+349}{5}-4\right)=0\)
\(\dfrac{x+329}{327}+\dfrac{x+329}{326}+\dfrac{x+329}{325}+\dfrac{x+329}{324}+\dfrac{x+329}{5}=0\)
\(\left(x+329\right)\left(\dfrac{1}{327}+\dfrac{1}{326}+\dfrac{1}{325}+\dfrac{1}{324}+\dfrac{1}{5}\right)=0\)
\(\Rightarrow\) x +329 =0 vay x= - 329
(\(\dfrac{x+2}{327}+1\)) +\(\left(\dfrac{x+3}{326}+1\right)+\left(\dfrac{x+4}{325}+1\right)+\left(\dfrac{x+5}{324}+1\right)+\left(\dfrac{x+349}{5}-4\right)=0\)