X x 12 = 92 845 + 96
X=
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\(\Leftrightarrow\left(\dfrac{x-11}{111}+1\right)+\left(\dfrac{x-12}{112}+1\right)=\left(\dfrac{x-23}{123}+1\right)+\left(\dfrac{x-24}{124}+1\right)\)
=>x+100=0
=>x=-100
a) 12 + ( 5 + x ) = 20
⇔ 5 + x = 8
⇔ x = 3
Vậy x = 3
b) 120 + ( 50 + x ) = 180
⇔ 50 + x = 60
⇔ x = 10
Vậy x = 10
c) 175 + ( 30 - x ) = 200
⇔ 30 - x = 25
⇔ x = 5
Vậy x = 5
d) 130 - ( 100 + x ) = 25
⇔ 100 + x = 105
⇔ x = 5
Vậy x = 5
e) 145 - ( 125 + x ) = 12
⇔ 125 + x = 133
⇔ x = 8
Vậy x = 8
j) 5 ( x + 12 ) + 22 = 92
⇔ 5x + 60 + 22 = 92
⇔ 5x + 82 = 92
⇔ 5x = 10
⇔ x = 2
Vậy x = 2
g) 3 ( x + 23 ) + 6 = 96
⇔ 3x + 69 + 6 = 96
⇔ 3x + 75 = 96
⇔ 3x = 21
⇔ x = 7
Vậy x = 7
Tìm x :
a) 12 + ( 5 + x ) = 20
5 + x =20 - 12
5 + x = 8
x = 3
b) 120 + ( 50 + x ) = 180
50 + x = 180-120
50 + x = 60
x = 60 - 50
x = 10
c) 175 + ( 30 - x ) = 200
30 -x = 200 - 175
30 - x = 25
x = 30-25
x=5
d) 130- ( 100 +x ) =25
100 + x =130 - 25
100+x= 105
x=105-100
x=5
e) 145 - (125+x) = 12
125+x = 145-12
125+x=133
x= 133-125
x=8
j) 5 (x+12) +22 = 92
5 ( x+12) = 92 -22
5 (x+12)=70
x+12=70 : 5
x + 12 = 14
x = 14-12
x=2
g) 2 (x+23) + 6 =96
2 (x+23) = 96 - 6
2 (x+23)= 90
x+23=90:2
x+23=45
x= 45-23
x=22
CHÚC BẠN HỌC TỐT
a) \(\Leftrightarrow\frac{x+2}{98}+\frac{x+4}{96}+2=\frac{x+6}{94}+\frac{x+8}{92}+2\)
\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{96}-\frac{x+100}{94}-\frac{x+100}{92}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{98}+\frac{1}{96}+\frac{1}{94}+\frac{1}{92}\right)=0\)
\(\Leftrightarrow x+100=0\Leftrightarrow x=-100\)
b)\(\Leftrightarrow\frac{x-12}{77}+\frac{x-11}{78}-2=\frac{x-74}{15}+\frac{x-73}{16}-2\)
\(\Leftrightarrow\frac{x-89}{77}+\frac{x-89}{78}-\frac{x-89}{15}-\frac{x-89}{16}=0\)
\(\Leftrightarrow\left(x-89\right)\left(\frac{1}{77}+\frac{1}{78}-\frac{1}{15}-\frac{1}{16}\right)=0\)
\(\Leftrightarrow x-89=0\Leftrightarrow x=89\)
Lời giải:
a)
PT \(\Leftrightarrow \frac{x+2}{98}+1+\frac{x+4}{96}+1=\frac{x+6}{94}+1+\frac{x+8}{92}+1\)
\(\Leftrightarrow \frac{x+100}{98}+\frac{x+100}{96}=\frac{x+100}{94}+\frac{x+100}{92}\)
\(\Leftrightarrow (x+100)\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)=0\)
Dễ thấy \(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}<0\) nên $x+100=0$
$\Rightarrow x=-100$
b)
PT \(\Leftrightarrow \frac{x-12}{77}-1+\frac{x-11}{78}-1=\frac{x-74}{15}-1+\frac{x-73}{16}-1\)
\(\Leftrightarrow \frac{x-89}{77}+\frac{x-89}{78}=\frac{x-89}{15}+\frac{x-89}{16}\)
\(\Leftrightarrow (x-89)\left(\frac{1}{77}+\frac{1}{78}-\frac{1}{15}-\frac{1}{16}\right)=0\)
Dễ thấy \(\frac{1}{77}+\frac{1}{78}-\frac{1}{15}-\frac{1}{16}< 0\)
\(\Rightarrow x-89=0\Rightarrow x=89\)
Ta có : \(\frac{x+2}{98}+\frac{x+4}{96}=\frac{x+6}{94}+\frac{x+2}{98}\)
\(\Rightarrow\frac{x+2}{98}+1+\frac{x+4}{96}+1=\frac{x+6}{94}+1+\frac{x+2}{98}+1\)
\(\Rightarrow\frac{x+100}{98}+\frac{x+100}{96}=\frac{x+100}{94}+\frac{x+100}{92}\)
=> \(\left(x+100\right)\left(\frac{1}{98}+\frac{1}{96}\right)=\left(x+100\right).\left(\frac{1}{94}+\frac{1}{92}\right)\)
=> \(\left(x+100\right).\left(\frac{1}{98}+\frac{1}{96}\right)-\left(x+100\right)\left(\frac{1}{94}+\frac{1}{92}\right)=0\)
=> \(\left(x+100\right).\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)=0\Rightarrow x+100=0\left(\text{vì }\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\ne0\right)\)
=> x = - 100
Vậy x = - 100
\(\frac{x+2}{98}+\frac{x+4}{96}=\frac{x+6}{94}+\frac{x+8}{92}\)
\(\Leftrightarrow\frac{x+2}{98}+1+\frac{x+4}{96}+1=\frac{x+6}{94}+1+\frac{x+8}{92}+1\)
\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{96}=\frac{x+100}{94}+\frac{x+100}{92}\)
\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{96}-\frac{x+100}{94}-\frac{x+100}{92}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)=0\)
Vì \(\frac{1}{92}>\frac{1}{94}>\frac{1}{96}>\frac{1}{98}\)nên \(\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)< 0\)
\(\Rightarrow x+100=0\Leftrightarrow x=-100\)
\(\Leftrightarrow\frac{x+2}{98}+\frac{x+4}{96}=\frac{x+6}{94}+\frac{x+8}{92}\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{96}=\frac{x+100}{94}+\frac{x+100}{92}\)
\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)=0.Ma:\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}< 0\Rightarrow x=-100\)
\(x^2+7x+12\)
\(=x^2+3x+4x+12\)
\(=x\left(x+3\right)+4\left(x+3\right)\)
\(=\left(x+3\right)\left(x+4\right)\)
\(a^{10}+a^5+1\)
\(=\left(a^{10}-a\right)+\left(a^5-a^2\right)+\left(a^2+a+1\right)\)
\(=a\left(a^9-1\right)+a^2\left(a^3-1\right)+\left(a^2+a+1\right)\)
\(=a\left(a^3-1\right)\left(a^3+1\right)+a^2\left(a^3-1\right)+\left(a^2+a+1\right)\)
\(=\left(a^4+a\right)\left(a^2+a+1\right)\left(a-1\right)+a^2\left(a-1\right)\left(a^2+a+1\right)+\left(a^2+a+1\right)\)
\(=\left(a^2+a+1\right)\left(a^5-a^4+a^2-a\right)+\left(a^3-a^2\right)\left(a^2+a+1\right)+\left(a^2+a+1\right)\)
\(=\left(a^2+a+1\right)\left(a^5-a^4+a^2-a+a^3-a^2+1\right)\)
\(=\left(a^2+a+1\right)\left(a^5-a^4+a^3-a+1\right)\)
\(x*12 = 92845+96 \)
\(x*12=92941\)
\(x = 92941:12\)
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