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3(2x+4)+2(x+1)=4x+94
=> 6x+12+2x+2-4x-94=0
=> 4x-80=0
=> 4x=80
=> x= 20
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\(\left(x+\dfrac{1}{5}\right)^2=\dfrac{9}{25}\)
\(\left(x+\dfrac{1}{5}\right)^2=\left(\dfrac{3}{5}\right)^2\)
➩ \(x+\dfrac{1}{5}=\dfrac{3}{5}\)
\(x=\dfrac{3}{5}-\dfrac{1}{5}\)
\(x=\dfrac{2}{5}\)
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\(100000-x=34796\)
\(x=100000-34796=65204\)
Còn lại sai đề hết rùi!
bạn thêm cái gì đó vào để giải đi mình cũng ko bít nữa làm ơn
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x2+4=x2 -4+8=(x-2)(x+2)+8 chia hết cho 2
(=) 8 chia hết cho x+2
(=) x+2 thuộc Ư(8)={ -8 ; -4 ; -2;-1;2;4;8)
(=)x ={-10;-6;-4;-3;-1;0;2;6}
vậy x....
#Học-tốt
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a)4.(x-3)=7.2-1.10
4.(x-3)=14-10
4.(x-3)=4
x-3=4:4
x-3=1
x=1+3
x=4
b)2.(x-51)=2.2.3+20
2.(x-51)=32
x-51=32:2
x-51=16
x=16+51
x=67
c)9.x-1=9
9.x=9+1
9.x=10
x=10:9
x=1,1111111...
(x-2).4=16
x-2=16:4
x-2=4
x=4+2
x=6
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a) \(\frac{x-1}{99}+\frac{x-2}{98}+\frac{x-3}{97}+\frac{x-4}{96}=4\)
\(\Rightarrow\frac{x-1}{99}-1+\frac{x-2}{98}-1+\frac{x-3}{97}-1+\frac{x-3}{96}-1=4-4\)
\(\Rightarrow\frac{x-100}{99}+\frac{x-100}{98}+\frac{x-100}{97}+\frac{x-100}{96}=0\)
\(\Rightarrow\left(x-100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\right)=0\)
\(\Rightarrow x-1=0\) ( vì \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\ne0\) )
Vậy x = 1
b) \(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}=3\)
\(\Rightarrow\frac{x+1}{99}+1+\frac{x+2}{98}+1+\frac{x+3}{97}+1=3-3\)
\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}=0\)
\(\Rightarrow\left(x+100\right).\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}\right)=0\)
Vì \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}\ne0\)
=> x + 100 = 0
=> x = -100
c) \(\frac{x-1}{99}+\frac{x-2}{49}+\frac{x-4}{32}=6\)
\(\Rightarrow\frac{x-1}{99}-1+\frac{x-2}{49}-2+\frac{x-4}{32}-3=6-6\)
\(\Rightarrow\frac{x-100}{99}+\frac{x-100}{49}+\frac{x-100}{32}=0\)
\(\Rightarrow\left(x-100\right)\left(\frac{1}{99}+\frac{1}{49}+\frac{1}{32}\right)=0\)
Vì \(\frac{1}{99}+\frac{1}{49}+\frac{1}{32}\ne0\)
=> x - 100 = 0
=> x = 100
Chúc bạn học tốt
có người khác trả lời trước rồi nên chị ko trả lời đâu nhé em trai
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Đặt B= 2 + 4 + 6 +........ + 2X
Tổng B có số số hạng theo x là: ( 2 x − 2 ) : 2 + 1 = x (2x−2):2+1=x (số)
Tổng B theo x là: ( 2 x + 2 ) ⋅ x : 2 = x 2 + x (2x+2)⋅x:2=x2+x
Thay B vào ta dc: x 2 + x = 210 x2+x=210 ⇔ x 2 + x − 210 = 0
⇔x2+x−210=0 ⇔ x 2 − 14 x + 15 x − 210 = 0
⇔x2−14x+15x−210=0 ⇔ x ( x − 14 ) + 15 ( x − 14 ) = 0 ⇔
x(x−14)+15(x−14)=0 ⇔ ( x + 15 ) ( x − 14 ) = 0 ⇔(x+15)(x−14)=0 ⇔ [ x = 14 ( t m ) x = − 15 ( l o a i )
(x - 2)x + 2 = (x - 2)x + 4 => (x - 2)x = (x - 2)x + 4 - 2 = (x - 2)x - 2 (vô lý)(loại). Vậy x ∊ ∅