a) x=1

vì 11*2.1chia hết cho 2.1-1

a ) x=2 

b ) x =9 

y = 12

a, 1/2 x 356 + 45/10 x 356 + 356 x 50/10

= 356 x ( 1/2 + 45/10 + 50/10 )

= 356 x 10

= 3560

b, x+4 + x+6 + x+8 + x+12 = 1290

(x+x+x+x) + (4+6+8+12) = 1290

4x + 30 = 1290

4x = 1290 - 30

4x = 1260

x = 1260 : 4

x = 315

 x+4+x+6+x+12=1290

x+ 4+6+12=1290

x+22=1290

x=1290-22

x=1268

Bạn viết như này ai hiểu

Viết lại đề lại sách nháp đi

\(\Rightarrow\frac{\left(\frac{518}{19}-\frac{342}{13}\right).\left(\frac{253}{236}-\frac{3}{318}\right)}{\left(\frac{3}{4}+x\right)\frac{27}{33}}\)=\(\frac{\frac{1}{3}.\frac{3}{36}+\frac{1}{3}.\frac{3}{14.17}}{\frac{1}{2}.\frac{2}{13.15}+\frac{1}{2}.\frac{2}{14.16}}\)

\(\frac{\frac{236}{2247}.\frac{247}{236}}{\left(\frac{3}{4}+x\right)\frac{27}{33}}\)=\(\frac{\frac{1}{3}.\left(\frac{1}{13}-\frac{1}{16}+\frac{1}{14}-\frac{1}{17}\right)}{\frac{1}{2}\left(\frac{1}{13}+\frac{1}{14}-\frac{1}{16}-\frac{1}{17}\right)}\)

\(\frac{1}{\left(\frac{3}{4}+x\right)\frac{27}{33}}\)=\(\frac{\frac{1}{3}}{\frac{1}{2}}\)

đến đây tự làm nhé cứ ra 13/12 là được

a) (x + 2)2 + 3(x + 1) ≥ x2 – 4

⇔ x2 + 4x + 4 + 3x + 3 ≥ x2 – 4

⇔ 7x + 7 ≥ –4

⇔ 7x ≥ –11

⇔ x ≥ -11/7

Tập nghiệm: S = {x|x ≥ -11/7}

b) 

⇔ 6(x – 1) – 4(x – 2) ≤ 12x – 3(x – 3)

⇔ 6x – 6 – 4x + 8 ≤ 12x – 3x + 9

⇔ 2x + 2 ≤ 9x + 9

⇔ –7x ≤ 7 ⇔ x ≥ –1

Tập nghiệm: S = {x|x ≥ -1}

\(x\left(x+1\right)=156\)

\(\Rightarrow x^2+x=156\)

\(\Rightarrow x^2+x-156=0\)

\(\Rightarrow x^2+13x-12x-156=0\)

\(\Rightarrow x\left(x+13\right)-12\left(x+13\right)=0\)

\(\Rightarrow\left(x+13\right)\left(x-12\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=12\\x=-13\end{matrix}\right.\)

___________________

\(x\left(x+1\right)=342\)

\(\Rightarrow x^2+x=342\)

\(\Rightarrow x^2+x-342=0\)

\(\Rightarrow x^2+19x-18x-342=0\)

\(\Rightarrow x\left(x+19\right)-18\left(x+19\right)=0\)

\(\Rightarrow\left(x+19\right)\left(x-18\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-19\\x=18\end{matrix}\right.\)

__________________

\(x\left(x+1\right)=650\)

\(\Rightarrow x^2+x=650\)

\(\Rightarrow x^2-x+650=0\)

\(\Rightarrow x^2+26x-25x-650=0\)

\(\Rightarrow x\left(x+26\right)-25\left(x+26\right)=0\)

\(\Rightarrow\left(x+26\right)\left(x-25\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-26\\x=25\end{matrix}\right.\)

______________________

\(x\left(x+1\right)=380\)

\(\Rightarrow x^2+x=380\)

\(\Rightarrow x^2+x-380=0\)

\(\Rightarrow x^2+20x-19x-380=0\)

\(\Rightarrow x\left(x+20\right)-19\left(x+20\right)=0\)

\(\Rightarrow\left(x+20\right)\left(x-19\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-20\\x=19\end{matrix}\right.\)

a, \(x\).(\(x\) + 1) = 156

156 = 2².3.13 = 12.13

Vậy \(x\).(\(x\) + 1) = 12.13

Vậy \(x\) = 12

b, \(x.\)(\(x\) + 1) =  342 

    342 = 2.3².19 = 18.19

    \(x\).(\(x+1\)) = 18.19

    \(x\) =   18

c, \(x\).(\(x\) + 1) = 650 

    650 = 2.5².13 = 25.26

    \(x\).(\(x\) +1) = 25.26

    \(x\) = 25

d, \(x\).(\(x\) +1) = 380 

    380 = 2².5.19 = 19.20

      \(x\).(\(x\) + 1) = 19.20

      \(x\)    = 19

Chọn B

C

Ủa, vậy đáp án nào đúng hơn? 

a) \(đk:\left\{{}\begin{matrix}x\ge0\\\sqrt{x}\ne2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)

b) \(x=3+2\sqrt{2}\Rightarrow\sqrt{x}=\sqrt{3+2\sqrt{2}}=\sqrt{\left(\sqrt{2}+1\right)^2}=\sqrt{2}+1\)

\(A=\dfrac{2\sqrt{x}-1}{\sqrt{x}-2}=\dfrac{2\left(\sqrt{2}+1\right)-1}{\sqrt{2}+1-2}=\dfrac{2\sqrt{2}+1}{\sqrt{2}-1}\)

c) \(A=\dfrac{2\sqrt{x}-1}{\sqrt{x}-2}=\dfrac{1}{2}\)

\(\Leftrightarrow4\sqrt{x}-2=\sqrt{x}-2\Leftrightarrow3\sqrt{x}=0\Leftrightarrow x=0\left(tm\right)\)

d) \(A=\dfrac{2\sqrt{x}-1}{\sqrt{x}-2}>2\)

\(\Leftrightarrow2\sqrt{x}-1>2\sqrt{x}-4\Leftrightarrow-1>-4\left(đúng\forall x\right)\)

e) \(A=\dfrac{2\sqrt{x}-1}{\sqrt{x}-2}=\dfrac{2\left(\sqrt{x}-2\right)}{\sqrt{x}-2}+\dfrac{3}{\sqrt{x}-2}=2+\dfrac{3}{\sqrt{x}-2}\in Z\)

\(\Rightarrow\sqrt{x}-2\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\)

Do \(x\ge0\)

\(\Rightarrow x\in\left\{1;9;25\right\}\)