2x + 7 chia hết cho x + 1

=> 2x + 2 + 5 chia hết cho x + 1

=> 2.(x + 1) + 5 chia hết cho x + 1

mà 2.(x + 1) chia hết cho x + 1

=> 5 chia hết cho x + 1

=> x + 1 thuộc Ư(5) = {-5; -1; 1; 5}

=> x thuộc {-6; -2; 0; 4}.

Ta có 2x+7 = 2x +2 +5 

Để 2x+7 chia hết cho x+1 thì 5 chia hết cho x+1 

=> x+1 thuộc Ư(5) = { -5 ; -1 ; 1 ;5 } 

Ta có bảng sau : 

 Vậy x = { -6 ; -2 ; 0 ; 4 } thì 2x+7 chia hết cho x+1

bạn qua wed khác wed này chỉ toán bạn nhé

co anh van nua

ĐKXĐ: x<>0

Ta có: \(8x^2+\frac{2}{x^2}\)

\(=2\left(4x^2+\frac{1}{x^2}\right)\)

\(=2\left\lbrack4x^2+\frac{1}{x^2}-2\cdot2x\cdot\frac{1}{x}+2\cdot2x\cdot\frac{1}{x}\right\rbrack=2\left\lbrack\left(2x-\frac{1}{x}\right)^2+4\right\rbrack\)

Ta có: \(8x^2+\frac{2}{x^2}-9\cdot\left|2x-\frac{1}{x}\right|-1=0\)

=>\(2\cdot\left\lbrack\left(\left|2x-\frac{1}{x}\right|^{}\right)^2+4\right\rbrack-9\cdot\left|2x-\frac{1}{x}\right|-1=0\)

=>\(2\cdot\left(\left|2x-\frac{1}{x}\right|\right)^2+8-9\cdot\left|2x-\frac{1}{x}\right|-1=0\)

=>\(2\cdot\left(\left|2x-\frac{1}{x}\right|\right)^2-9\cdot\left|2x-\frac{1}{x}\right|+7=0\)

=>\(\left(2\cdot\left|2x-\frac{1}{x}\right|-7\right)\left(\left|2x-\frac{1}{x}\right|-1\right)=0\)

TH1: \(2\cdot\left|2x-\frac{1}{x}\right|-7=0\)

=>\(\left|2x-\frac{1}{x}\right|=\frac72\)

=>\(\left[\begin{array}{l}2x-\frac{1}{x}=\frac72\\ 2x-\frac{1}{x}=-\frac72\end{array}\right.\Rightarrow\left[\begin{array}{l}\frac{2x^2-1}{x}=\frac72\\ \frac{2x^2-1}{x}=-\frac72\end{array}\right.\Rightarrow\left[\begin{array}{l}4x^2-2=7x\\ 4x^2-2=-7x\end{array}\right.\)

=>\(\left[\begin{array}{l}4x^2-7x-2=0\\ 4x^2+7x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}4x^2-8x+x-2=0\\ 4x^2+8x-x-2=0\end{array}\right.\)

=>\(\left[\begin{array}{l}\left(x-2\right)\left(4x+1\right)=0\\ \left(x+2\right)\left(4x-1\right)=0\end{array}\right.\)

=>x∈{2;-2;-1/4;1/4}

TH2: \(\left|2x-\frac{1}{x}\right|-1=0\)

=>\(\left|2x-\frac{1}{x}\right|=1\)

=>\(\left[\begin{array}{l}2x-\frac{1}{x}=1\\ 2x-\frac{1}{x}=-1\end{array}\right.\Rightarrow\left[\begin{array}{l}\frac{2x^2-1}{x}=1\\ \frac{2x^2-1}{x}=-1\end{array}\right.\Rightarrow\left[\begin{array}{l}2x^2-1-x=0\\ 2x^2+x-1=0\end{array}\right.\)

=>\(\left[\begin{array}{l}2x^2-2x+x-1=0\\ 2x^2+2x-x-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}\left(x-1\right)\left(2x+1\right)=0\\ \left(x+1\right)\left(2x-1\right)=0\end{array}\right.\)

=>x∈{1;-1;-1/2;1/2}

Câu 1: C

Câu 2: B

Câu 3: A

B1:

a) mdd = 80 + 120 = 200 (g)

C%Na2CO3 = 80/200 = 40%

b) C%K3PO4 = 63,6/200 = 31,8%

B2:

a) mBaCl2 = 15% . 200 = 30 (g)

b) mKNO3 = 12% . 250 = 30 (g)

B3: 

a) mNaCl = 250 . 11,7% = 29,25 (g)

nNaCl = 29,25/58,5 = 0,5 (mol)

b) nKMnO4 = 0,25 . 4 = 1 (mol)

B4:

a) mFe(NO3)3 = 0,25 . 242 = 60,5 (g)

mdd = 189,5 + 60,5 = 250 (g)

C%Fe(NO3)3 = 60,5/250 = 24,2%

b) nK3PO4 = 42,4/212 = 0,2 (mol)

CMK3PO4 = 0,2/0,5 = 0,4M

B5:

a) nBa(OH)2 = 0,3 . 1,5 = 0,45 (mol)

mBa(OH)2 = 0,45 . 171 = 76,95 (g)

b) nNaOH = 6/40 = 0,15 (mol)

VddNaOH = 0,15/2 = 0,075 (l)

Nhìn thôi đã thấy ngán, em nên tách ra

\(x\times4+\frac{1}{2}\times x=55,35\)

\(\Leftrightarrow x\times\left(4+\frac{1}{2}\right)=55,35\)

\(\Leftrightarrow x\times4,5=55,35\)

\(\Leftrightarrow x=55,35:4,5\)

\(\Leftrightarrow x=12,3\)

Vậy x= 12,3

x.4+1/2.x=55,35
x.(4+1/2)=55.35
x.9/2=55,35
x=55,35:9/2
x=12,3
vì là máy ko có dấu nhân nên thay dấu nhân là dấu chấm

Chọn C