Tìm x biết 67 - x 76 = 7 19
A. 29
B. 39
C. 49
D. 59
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Lí do = 0 nè:
\(\frac{x-39}{19}+\frac{x-49}{29}=\frac{x-5}{39}+\frac{x-69}{49}\)
=> \(\frac{x-39}{19}+\frac{x-49}{29}=0+\left(\frac{x-5}{39}+\frac{x-69}{49}\right)\)
=> \(\frac{x-39}{19}+\frac{x-49}{29}-\left(\frac{x-5}{39}+\frac{x-69}{49}\right)=0\)
=> \(\frac{x-39}{19}+\frac{x-49}{29}-\frac{x-5}{39}-\frac{x-69}{49}=0\)
Câu 2:
a: x-158=32
=>x=158+32
=>x=190
b: \(x\cdot24=264\)
=>\(x=\dfrac{264}{24}\)
=>x=11
c: \(6x+9=3^7:3^4\)
=>\(6x+9=3^3\)
=>6x+9=27
=>6x=18
=>x=18/6=3
Câu 1:
a: \(86\cdot19+14\cdot19\)
\(=19\left(86+14\right)\)
\(=19\cdot100=1900\)
b: \(4\cdot\left(-5\right)^2-104\cdot\left(-5\right)^2\)
\(=4\cdot25-104\cdot25\)
\(=25\left(4-104\right)=-100\cdot25=-2500\)
c: \(7\cdot\left(-2\right)\cdot8\left(-5\right)\)
\(=7\cdot2\cdot8\cdot5\)
\(=56\cdot10=560\)
d: \(59-\left[59+\left(-76\right)\right]\)
\(=59-59+76\)
=76
a,x/8=2/x
=>x2=16
=>x=-4;4
vậy x=-4;4
b,\(\frac{x-39}{19}+\frac{x-49}{29}=\frac{x-59}{39}+\frac{x-69}{49}\)
\(\Rightarrow\left(\frac{x-39}{19}+1\right)+\left(\frac{x-49}{29}+1\right)=\left(\frac{x-59}{39}+1\right)+\left(\frac{x-69}{49}+1\right)\)
\(\Rightarrow\frac{x-20}{19}+\frac{x-20}{29}=\frac{x-20}{39}+\frac{x-20}{49}\)
\(\Rightarrow\frac{x-20}{19}+\frac{x-20}{29}-\frac{x-20}{39}-\frac{x-20}{49}=0\)
\(\Rightarrow\left(x-20\right)\left(\frac{1}{19}+\frac{1}{29}-\frac{1}{39}-\frac{1}{49}\right)=0\)
vì \(\frac{1}{19}+\frac{1}{29}-\frac{1}{39}-\frac{1}{49}\ne0\Rightarrow x-20=0\Rightarrow x=20\)
vậy x=20
\(\frac{x}{8}=\frac{2}{x}\Rightarrow x^2=8.2\)
\(x^2=16\Rightarrow x^2=\left(\pm4\right)^2\Rightarrow x=4;x=-4\)
Học tốt
lấy chữ số tận cùng ta có: 9x9=81 mà 1 x 1=1
mà 6 x 6 x6x6x6 = có chữ số tận cùng là 6
mà 1 x 6 =6
vậy có chữ số tận cùng là 6
1) 39×51+39×49=39x(51+49)=39x100=3900
2. phân tích 100 thành 50 r lm như câu 1
1. = 39 x (51 + 49)
= 39 x 100
= 3900
2. = 50 x 2 x 31 + 100 x 19
= 100 x 31 + 100 x 19
= 100 x (31 + 19)
= 100 x 50
= 5000
\(\frac{x-39}{19}+\frac{x-49}{29}=\frac{x-59}{39}+\frac{x-69}{49}\)
Công mỗi vế cho 2 ta được:
\(\frac{x-39}{19}+\frac{x-49}{29}+2=\frac{x-59}{39}+\frac{x-69}{49}+2\)
\(\frac{x-39}{19}+1+\frac{x-49}{29}+1=\frac{x-59}{39}+1+\frac{x-69}{49}+1\)
\(\frac{x-39}{19}+\frac{19}{19}+\frac{x-49}{29}+\frac{29}{29}=\frac{x-59}{39}+\frac{39}{39}+\frac{x-69}{49}+\frac{49}{49}\)
\(\frac{x-20}{19}+\frac{x-20}{29}=\frac{x-20}{39}+\frac{x-20}{49}\)
\(\frac{x-20}{19}+\frac{x-20}{29}-\frac{x-20}{39}-\frac{x-20}{49}=0\)
\(\left(x-20\right).\frac{1}{19}+\left(x-20\right).\frac{1}{29}-\left(x-20\right).\frac{1}{39}-\left(x-20\right).\frac{1}{49}=0\)
Đặt thừa số chung (x-20) ra ngoài
\(\left(x-20\right).\left(\frac{1}{19}+\frac{1}{29}-\frac{1}{39}-\frac{1}{49}\right)=0\)
\(\text{Vì }\frac{1}{19}+\frac{1}{29}-\frac{1}{39}-\frac{1}{49}\ne0\text{ nên }x-20=0\Rightarrow x=20\)
Đáp án là B