Sửa đề: \(\dfrac{74-x}{26}+\dfrac{75-x}{25}+\dfrac{76-x}{24}+\dfrac{77-x}{23}+\dfrac{78-x}{22}=-5\)Ta có: \(\dfrac{74-x}{26}+\dfrac{75-x}{25}+\dfrac{76-x}{24}+\dfrac{77-x}{23}+\dfrac{78-x}{22}=-5\)

\(\Leftrightarrow\dfrac{74-x}{26}+1+\dfrac{75-x}{25}+1+\dfrac{76-x}{24}+1+\dfrac{77-x}{23}+1+\dfrac{78-x}{22}+1=0\)

\(\Leftrightarrow\dfrac{100-x}{26}+\dfrac{100-x}{25}+\dfrac{100-x}{24}+\dfrac{100-x}{23}+\dfrac{100-x}{22}=0\)

\(\Leftrightarrow\left(100-x\right)\left(\dfrac{1}{26}+\dfrac{1}{25}+\dfrac{1}{24}+\dfrac{1}{23}+\dfrac{1}{22}\right)=0\)

mà \(\dfrac{1}{26}+\dfrac{1}{25}+\dfrac{1}{24}+\dfrac{1}{23}+\dfrac{1}{22}>0\)

nên 100-x=0

hay x=100

Vậy: S={100}

Ta có : \(\dfrac{74-x}{26}+\dfrac{75-x}{25}+\dfrac{76-x}{24}+\dfrac{77-x}{23}+\dfrac{78-x}{22}=-5\)

\(\Leftrightarrow\dfrac{74-x}{26}+\dfrac{75-x}{25}+\dfrac{76-x}{24}+\dfrac{77-x}{23}+\dfrac{78-x}{22}+5=0\)

\(\Leftrightarrow\dfrac{74-x}{26}+1+\dfrac{75-x}{25}+1+\dfrac{76-x}{24}+1+\dfrac{77-x}{23}+1+\dfrac{78-x}{22}+1=0\)

\(\Leftrightarrow\dfrac{100-x}{26}+\dfrac{100-x}{25}+\dfrac{100-x}{24}+\dfrac{100-x}{23}+\dfrac{100-x}{22}=0\)

\(\Leftrightarrow\left(100-x\right)\left(\dfrac{1}{26}+\dfrac{1}{25}+\dfrac{1}{24}+\dfrac{1}{23}+\dfrac{1}{22}\right)=0\)

Thấy : \(\dfrac{1}{26}+\dfrac{1}{25}+\dfrac{1}{24}+\dfrac{1}{23}+\dfrac{1}{22}\ne0\)

\(\Rightarrow100-x=0\)

\(\Leftrightarrow x=100\)

Vậy ...

a)    \(\left(-25\right).5.\left(-4\right).2\)

\(=\left[\left(-25\right).\left(-4\right)\right].\left(2.5\right)\)

\(=100.10=1000\)

b)   \(\left(-456\right)-\left(67-456\right)\)

\(=-456-67+456\)

\(=\left(456-456\right)-67\)

\(=67\)

a) ((-25).(-4)).5.2

=100.10

=1000

b) (-456)-67+456

=(-456)+(-67)+456

=0+-(-67)

=-67

c)78.45+-145.78

=78.(45+145)

=78.190

=Tự làm

Đáp án cần chọn là: D

a) 5 + (-5) = 0

b) 71-52 = 19

c) 28 + (-82) = -64

d) 47 + (-30) = 17

e) (-41) + (-59) = -100

f) 75 + (-10) = 65

t) 26 - 43 = -17

y) 90 - 25 = 650

o) 72 + (-9) = 63

p) (-3) + 3 = 0

\(a;\frac{2^{78}+2^{79}+2^{80}}{2^{77}+2^{76}+2^{75}}=\frac{2^{78}\left(1+2+2^2\right)}{2^{75}\left(1+2+2^2\right)}=2^3=8\)

b) \(\frac{5^{56}+5^7}{5^{49}+1}=\frac{5^7\left(5^{49}+1\right)}{5^{49}+1}=5^7\)

a: 1152-(374+1152)+(-65+374)

=1152-374-1152-65+374

=(1152-1152)+(-374+374)-65

=0+0-65

=-65

b: \(\left\lbrack53+\left(-76\right)\right\rbrack-\left\lbrack-76-\left(-53\right)\right\rbrack\)

\(=\left\lbrack53-76\right\rbrack-\left\lbrack-76+53\right\rbrack\)

=0

c: \(-48+48\cdot\left(-78\right)+21\cdot\left(-48\right)\)

\(=\left(-48\right)\left(1+78+21\right)\)

\(=-48\cdot100=-4800\)

d:Sửa đề: \(\left(-12\right)\cdot47+\left(-12\right)\cdot52+\left(-12\right)\)

\(=\left(-12\right)\left(47+52+1\right)\)

\(=-12\cdot100=-1200\)

e: \(\left(-1\right)+\left(-3\right)+\cdots+\left(-199\right)+\left(-201\right)\)

=-(1+3+...+199+201)

Số số hạng của dãy số 1;3;...;199;201 là:

\(\frac{201-1}{2}+1=\frac{200}{2}+1=100+1=101\) (số)

Tổng của dãy số 1;3;...;199;201 là:

\(\left(201+1\right)\cdot\frac{101}{2}=202\cdot\frac{101}{2}=101\cdot101=10201\)

=>(-1)+(-3)+...+(-199)+(-201)=-10201

f:

Số số hạng hạng trong dãy số 17;20;23;..;53;56 là:

(56-17):3+1=39:3+1=13+1=14(số)

17+(-20)+23+(-26)+...+53+(-56)

=(17-20)+(23-26)+...+(53-56)

=(-3)+(-3)+...+(-3)

\(=-3\cdot\frac{14}{2}=-21\)

\(\frac{74-x}{26}+\frac{75-x}{25}+\frac{76-x}{24}+\frac{77-x}{23}+\frac{78-x}{22}=-5\)

\(\frac{74-x}{26}+1+\frac{75-x}{25}+1+\frac{76-x}{24}+1+\frac{77-x}{23}+1+\frac{78-x}{22}=-5+5\)

\(\frac{74-x}{26}+\frac{26}{26}+\frac{75-x}{25}+\frac{25}{25}+\frac{76-x}{24}+\frac{24}{24}+\frac{77-x}{23}+\frac{23}{23}+\frac{78-x}{22}+\frac{22}{22}=0\)

\(\frac{100-x}{26}+\frac{100-x}{25}+\frac{100-x}{24}+\frac{100-x}{23}+\frac{100-x}{22}=0\)

\(\left(100-x\right)\left(\frac{1}{26}+\frac{1}{25}+\frac{1}{24}+\frac{1}{23}+\frac{1}{22}\right)=0\)

=>100-x=0    ( \(\left(\frac{1}{26}+\frac{1}{25}+\frac{1}{24}+\frac{1}{23}+\frac{1}{22}\right)\ne0\))

x=100

hahaha

Cộng 1 vào mỗi hạng tử trong vế trái là dc