\(\left(x^2+1\right)\left(x-1\right)=0\)

Ta có: \(x^2\ge0\forall x\)

\(\Rightarrow x^2+1\ge1>0\)

\(\Rightarrow x-1=0\)

\(\Rightarrow x=1\)

(x²+1).(x-1)=0

=> có 2 TH : (x²+1) = 0 hoặc (x-1) = 0

TH1 : x²+1 = 0

          x²      = 0 - 1

          x²      = -1

=> Không tìm được x hợp lí

=> TH 1 (loại) 

TH2 : x-1=0

          x   = 0+1

          x   = 1

Vậy x  = 1

\(1]\left(x+5\right)\left(x-4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+5=0\\x-4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-5\\x=4\end{matrix}\right.\)

\(2]\) (Tôi giải ở câu hỏi khác của bạn rồi nhé.)

\(3]25-\left(3+x\right)^2=3.\left(-2\right)^3\)

\(\Rightarrow25-\left(3+x\right)^2=-24\)

\(\Rightarrow25-\left(3+x\right)^2+24=0\)

\(\Rightarrow7^2-\left(3+x\right)^2=0\)

\(\Rightarrow\left[7+\left(3+x\right)\right]\left[7-\left(3+x\right)\right]=0\)

\(\Rightarrow\left(x+10\right)\left(-x+4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+10=0\\-x+4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-10\\x=4\end{matrix}\right.\)

\(4]\left(x+1\right)^3+9=-116\)

\(\Rightarrow\left(x+1\right)^3=-125\)

\(\Rightarrow\left(x+1\right)^3=\left(-5\right)^3\)

\(\Rightarrow x+1=-5\)

\(\Rightarrow x=-6\)

(X+5)(x-4) = 0

x + 5 = 0 hặc x - 4 = 0

X + 5 = 0 ⇒ X = -5;    x -4 = 0 ⇒ X = 4

x ϵ \(\left\{-5;4\right\}\)

2; (x² + 1)(x-1) = 0

x -1 = 0 ⇒ x = 1; x ϵ \(\left\{1\right\}\)

3; 25 - (3 + x)² = 3.(-2)³

     (3+x)² = 25 + 3x8

      (3+ x)² = 49

       3 + x = \(\mp\)7

     3 + x = 7 ⇒ x = 7-3 =4

      3 + x = -7 ⇒ x =-10

            x ϵ \(\left\{-10;4\right\}\)

4; (x+1)³  + 9 = - 116

(x + 1)³ = -9 -116

(x + 1)³ = - 125 = (-5)³

X + 1 = -5

X = -1  - 5 = -6 

x ϵ \(\left\{-6\right\}\)

giúp mình với

\(A=\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{97.99}\)

\(=\dfrac{1}{2}.\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{97.99}\right)\)

\(=\dfrac{1}{2}.\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)

\(=\dfrac{1}{2}.\left(\dfrac{1}{3}-\dfrac{1}{99}\right)\)

\(=\dfrac{16}{99}\)

___

\(B=\dfrac{3}{3.5}+\dfrac{3}{5.7}+...+\dfrac{3}{97.99}\)

\(=\dfrac{3}{2}.\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{97.99}\right)\)

\(=\dfrac{3}{2}.\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)

\(=\dfrac{3}{2}.\left(\dfrac{1}{3}-\dfrac{1}{99}\right)\)

\(=\dfrac{16}{33}\)

___

\(M=1+3+5+7+...+9\)

Số số hạng: \(\left(99-1\right):2+1=50\) số hạng

Tổng: \(\left(99+1\right).50:2=2500\)

___

\(C=3+7+11+...+103\)

Số số hạng: \(\left(103-3\right):4+1=26\)

Tổng: \(\left(103+3\right).26:2=1378\)

x=0

x=0

`1,` Thay `a=-5;b=7;c=-12` vào `A` có:

`A=-5-7-12=-24`

`2,` Thay `a=-2;b=4` vào `B` có:

`B=(-2)^2-5+2.(-2).4-4^2=4-5-16-16=-33`

`3,` Thay `x=1;y=3` vào `C` có:

`C=-7.1+3.1:6-(1+3)=-7+1/2-4=-21/2`

`4,` Thay `x=-10` vào`D` có:

`D=-7.(-10)+(-10):(-5)-20=70+2-20=52`

1,A=a-b+c

=> A=-5-7+(-12)

=> A=-24

2, B= a2 -5+2ab-b2-

=> B=(-2)^2.4-(4)^2

=> B=-9

3, C= -7x+3x:6-(x+y)

=> C=-7.1+3.1:6-(1+4)

=>C=-115

4, D= -7x+x:(-5)-20

=>D= -7.(-10)+(-10):(-5)-20

=>D=52

ĐỀ BÀI LÀ SAO BẠN ?

0,75 x 4 - X = 0,25 : 1/8

3 - X = 2

X = 3-2

X = 1

TL:
\(0,75\times4-x=0,25:\dfrac{1}{8}\)
\(\Rightarrow3-x=0,25:0,125\)
\(\Rightarrow3-x=2 \)
\(\Rightarrow x=1\)
Vậy x=1
HT!

\(S=7.\left(1+\dfrac{1}{3}+\dfrac{1}{6}+...+\dfrac{1}{55}\right)\)

\(=7.\left(\dfrac{2}{2}+\dfrac{2}{6}+\dfrac{2}{12}+...+\dfrac{2}{110}\right)\)

\(=14.\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{110}\right)\)

\(=14.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{10}-\dfrac{1}{11}\right)\)

\(=14.\dfrac{10}{11}=\dfrac{140}{11}.\)

\(S=7\left(1+\dfrac{1}{3}+\dfrac{1}{6}+...+\dfrac{1}{55}\right)\)

\(S=14\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+..+\dfrac{1}{110}\right)\)

\(S=14\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{10\cdot11}\right)\)

 \(S=14\)\(\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+..+\dfrac{1}{10}-\dfrac{1}{11}\right)\)

S\(=14\left(\dfrac{1}{1}-\dfrac{1}{11}\right)=14\cdot\dfrac{10}{11}=\dfrac{140}{11}\)