3/x+2 = 5/2x+1
Tim x
Bai 1tim x
A,123-5.(x+4)=38
B,10+2x=2.(32 -1)
a,123-5(x+4)=28
(=)5(x+4)=85
(=) x+4=17
(=) x=13
b,10+2x=2.\(\left(3^2-1\right)\)
(=) 10+2x=16
(=) 2x = 6
(=)x=3
A, 123-5.(x+4)=38
5.(x+4)=123-38
5.(x+4) =85
(x+4)=85:5
(x+4)=17
x =17-4
x =13
B, 10+2x=2(\(3^2-1\))
10+2x=2.8
10+2x=16
2x=16-10
2x=6
x=6:2
x =3
a, 123 - 5(x + 4) = 38
=>5(x + 4) = 85
=> x + 4 = 17
=> x = 13
vậy_
10 + 2x = 2(32 - 1)
=> 10 + 2x = 2.8
=> 10 + 2x = 16
=> 2x = 6
=> x = 3
vậy_
BT:
1Tim x
a)(x+2)(x+3)-(x-2)(x+5)=6
b)3(1-4x)(x-1)+4(3x-2)(x+3)=-27
c)5(2x+3)(y+2)-2(5y-4)(y-1)=75
d)(3x+2)(2x+9)-(x+2)(6x+1)=(x+1)-(x-6)
a: \(\Leftrightarrow x^2+5x+6-\left(x^2+3x-10\right)=6\)
\(\Leftrightarrow x^2+5x+6-x^2-3x+10=6\)
=>2x+16=6
=>2x=-10
hay x=-5
b: \(\Leftrightarrow3\left(x-1-4x^2+4x\right)+4\left(3x^2+9x-2x-6\right)=-27\)
\(\Leftrightarrow3\left(-4x^2+5x-1\right)+4\left(3x^2+7x-6\right)=-27\)
\(\Leftrightarrow-12x^2+15x-3+12x^2+28x-24=-27\)
=>43x=0
hay x=0
c: \(\Leftrightarrow5\left(2y^2+4y+3y+6\right)-2\left(5y^2-5y-4y+4\right)=75\)
\(\Leftrightarrow10y^2+35y+30-10y^2+18y-8=75\)
=>53y=53
hay y=1
d: \(\Leftrightarrow6x^2+27x+4x+18-\left(6x^2+x+12x+2\right)=x+1-x+6=7\)
\(\Leftrightarrow6x^2+31x+18-6x^2-13x-2=7\)
=>18x+16=7
=>18x=-9
hay x=-1/2
1tim x ∈ Z thoa man
a)/x/+/-5/=/-37/ b)/-6/./x/=/54/
c)/x/>21 d)/x/<-1
2tim x
a)x+3=/-3/+/-7/ b)-2</x/<2
a) |x| + |-5| = |-37|
<=> |x| + 5 = 37
<=> |x| = 37 - 5 = 32
=> x \(\in\) {32 ; -32}
b)|-6| . |x| = |54|
<=> 6 . |x| = 54
|x| = 54 : 6 = 9
=> x \(\in\){9;-9}
c) |x| > 21
Có |x| \(\ge\) 0 > 21
=> |x| \(\in\) { 22 ; 23 ; 24 ; 25 ; ....}
=> x \(\in\) { 22; -22 ; 23; -23; 24; -24; 25; -25; ....}
1tim các số nguyên
a,x-2*y+3=7
b,x.(y-5)=-9
c,x-7.x-3=5
d,xy+3x-7y=21
e,xy+7x-2y=11
thử thach đi nao ae !
Bai 1Tim x
a)(x+1)+(x+2)+(x+3)+...+(x+28)=462
b)890:x=35 dư 15
|x+3|>1tim x
1Tim x
a, /x+5/ - x + 23 = 27
b, 500 . x + 78 .x = 1056
a) \(\left|x+5\right|-x+23=27\)
\(\Leftrightarrow\left|x+5\right|-x+23-27=0\)
\(\Leftrightarrow\left|x+5\right|-x+\left(23-27\right)=0\)
\(\Leftrightarrow\left|x+5\right|-x+\left(-4\right)=0\)
\(\Leftrightarrow\left|x+5\right|-x-4=0\)
\(\Leftrightarrow\left|x+5\right|-\left(x+4\right)=0\)
\(\Leftrightarrow\left|x+5\right|=x+4\)
Xét trường hợp 1:
\(x+5=x+4\)
\(\Leftrightarrow x+5-\left(x+4\right)=0\)
\(\Leftrightarrow x+5-x-4=0\)
\(\Leftrightarrow1=0\left(loai\right)\)
Xét trường hợp 2:
\(x+5=-\left(x+4\right)\)
\(\Leftrightarrow x+5-\left[-\left(x+4\right)\right]=0\)
\(\Leftrightarrow x+5+x+4=0\)
\(\Leftrightarrow2x+9=0\)
\(\Leftrightarrow2x=-9\)
\(\Leftrightarrow x=\dfrac{-9}{2}\left(thoa\right)\)
Vậy \(x=\dfrac{-9}{2}\)
bài 1tim x,y,z biết
a) 4x=3y, 5y=3z và 2x-3y+z=6
b)\(\frac{x}{y}=\frac{3}{4},\frac{y}{z}=\frac{5}{7}\)và 2x+3y-z=186
c)\(\frac{6}{11}.x=\frac{9}{2}.y=\frac{18}{5}.z\) và x-y+=-196
d)2x=3y=5z và tri tuyệt đối của x+y-z=95
e)\(\frac{x-1}{2}=\frac{y+3}{4}=\frac{z-5}{6}\) và 5z-3x-4y=50
f)\(\frac{4}{x+1}=\frac{2}{x-2}=\frac{3}{z+2}\) và xyz=12
bài 2
a) cmr:\(3^{x+1}+3^{x+2}+3^{x+3}+.....+3^{x+100}\) chia hết cho 120
1Tim x biet
\(\frac{x+1}{9}+\frac{x+2}{8}=\frac{x+3}{7}+\frac{x+4}{6}\)
\(\frac{x+1}{9}+\frac{x+2}{8}=\frac{x+3}{7}+\frac{x+4}{6}\)
\(\Rightarrow\frac{x+1}{9}+1+\frac{x+2}{8}+1=\frac{x+3}{7}+1+\frac{x+4}{6}+1\)
\(\Rightarrow\frac{x+10}{9}+\frac{x+10}{8}=\frac{x+10}{7}+\frac{x+10}{6}\)
\(\Rightarrow\frac{x+10}{9}+\frac{x+10}{8}-\frac{x+10}{7}-\frac{x+10}{6}=0\)
\(\Rightarrow\left(x+10\right)\left(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)=0\)
\(\Rightarrow x+10=0\) vì \(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\ne0\)
\(\Rightarrow x=-10\)
=>\(\frac{x+1}{9}\)+1+\(\frac{x+2}{8}\)+1=\(\frac{x+3}{7}\)+1+\(\frac{x+4}{6}\)+1
=>\(\frac{x+10}{9}\)+ \(\frac{x+10}{8}\)- \(\frac{x+10}{7}\)- \(\frac{x+10}{6}\)= 0
=>(x+10)x(\(\frac{1}{9}\)+ \(\frac{1}{8}\)+ \(\frac{1}{7}\)+\(\frac{1}{6}\))=0
=>x+10=0
=>x= -10
\(\frac{x+1}{9}+\frac{x+2}{8}=\frac{x+3}{7}+\frac{x+4}{6}\)
\(\frac{x+1}{9}+1+\frac{x+2}{8}+1=\frac{x+3}{7}+1+\frac{x+4}{6}+1\)
\(\frac{x+10}{9}+\frac{x+10}{8}=\frac{x+10}{7}+\frac{x+10}{6}\)
\(\frac{x+10}{9}+\frac{x+10}{8}-\frac{x+10}{7}-\frac{x+10}{6}=0\)
\(\left(x+10\right)\left(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)=0\)
Vì \(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\ne0\)nên x +10 = 0
x = -10.