Lớp 8 chưa học Delta, bạn nên phân tích ra rồi tìm giá trị của x nhé.

`#3107.101107`

\(\left(x^2-9\right)-9\left(x-3\right)^2=0\\ \Rightarrow\left(x-3\right)\left(x+3\right)-9\left(x-3\right)^2=0\\ \Rightarrow\left(x-3\right)\left[x+3-9\left(x-3\right)\right]=0\\ \Rightarrow\left[{}\begin{matrix}x-3=0\\x+3-9x+27=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\-8x+30=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\-8x=-30\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\8x=30\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{15}{4}\end{matrix}\right.\)

Vậy, \(x\in\left\{3;\dfrac{15}{4}\right\}.\)

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Các HĐT sử dụng trong bài:

\(\left(A-B\right)^2=A^2-2AB+B^2\\ A^2-B^2=\left(A-B\right)\left(A+B\right).\)

`15.`

n của \(\text{KClO}_3\) có trong phản ứng:

\(\text{n}_{\text{KClO}_3}=\dfrac{\text{m}_{\text{KClO}_3}}{\text{M}_{\text{KClO}_3}}=\dfrac{6,125}{119}\approx0,05\left(\text{mol}\right)\)

PTHH: \(\text{2KClO}_3\text{ }\underrightarrow{\text{ }\text{ t}^0\text{ }\text{ }}2\text{KCl}+3\text{O}_2\)

\(2\text{n}_{\text{KClO}_3}=3\text{n}_{\text{O}_2}\)

\(\Rightarrow\text{n}_{\text{O}_2}=0,075\text{ mol}\)

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PTHH: \(\text{Zn + O}_2\rightarrow\text{ZnO}_2\)

\(\text{n}_{\text{O}_2}=\text{n}_{\text{ZnO}_2}=0,075\text{ mol}\)

m của \(\text{ZnO}_2\) sau phản ứng:

\(\text{m}_{\text{ZnO}_2}=\text{n}_{\text{ZnO}_2}\cdot\text{M}_{\text{ZnO}_2}=0,075\cdot97=7,275\left(\text{g}\right)\)

`#3107.101107`

`13.`

n của Cl₂:

\(\text{V}_{ \text{Cl}_2}=\text{n}_{\text{Cl}_2}\cdot24,79\\ \text{n}_{\text{Cl}_2}=\dfrac{3,7185}{24,79}=0,15\left(\text{mol}\right)\)

PTHH: \(\text{Zn + Cl}_2\rightarrow\text{ZnCl}_2\)

Theo pt: \(\text{n}_{\text{Cl}_2}=\text{n}_{\text{ZnCl}_2}=0,15\text{ mol}\)

m của ZnCl₂ sau phản ứng:

\(\text{ m}_{\text{ZnCl}_2}=\text{n}_{\text{ZnCl}_2}\cdot\text{M}_{\text{ ZnCl}_2}=0,15\cdot136=20,4\left(\text{g}\right)\)

`14.`

Số nguyên tử H có trong phân tử nước:

\(2\cdot6\cdot10^{23}=12\cdot10^{23}\) (nguyên tử)

`#3107.101107`

`( 12 + 77 +34 + 23 + 88 + 45 + 55 + 66 ) \times ( 45 200 - 1808 \times 25 )`

`= [(12 + 88) + (45 + 55) + (77 + 23) + (34 + 66)] \times (45200 - 45200)`

`= (100 + 100 + 100 + 100) \times 0`

`= 0`

`#3107.101107`

`8)`

\(15\cdot24-14\cdot5\left(145\right)\div5-27\\ =15\cdot24-14\cdot145-27\\ =360-2030-27\\ =-1697\)

`14.`

`a)`

\(x-\dfrac{1}{2}=-\dfrac{2}{3}\\ \Rightarrow x=-\dfrac{2}{3}+\dfrac{1}{2}\\ \Rightarrow x=-\dfrac{1}{6}\)

Vậy, \(x=-\dfrac{1}{6}\)

`b)`

\(2x+\dfrac{1}{5}=\dfrac{1}{2}\\ \Rightarrow2x=\dfrac{1}{2}-\dfrac{1}{5}\\ \Rightarrow2x=\dfrac{3}{10}\\ \Rightarrow x=\dfrac{3}{10}\div2\\ \Rightarrow x=\dfrac{3}{20}\)

Vậy, \(x=\dfrac{3}{20}\)

`c)`

\(\left(-\dfrac{5}{9}\right)^{10}\div x=\left(-\dfrac{5}{9}\right)^8\\ \Rightarrow x=\left(-\dfrac{5}{9}\right)^{10}\div\left(-\dfrac{5}{9}\right)^8\\ \Rightarrow x=\left(-\dfrac{5}{9}\right)^{10-8}\\ \Rightarrow x=\left(-\dfrac{5}{9}\right)^2\\ \Rightarrow x=\dfrac{25}{81}\)

Vậy, \(x=\dfrac{25}{81}\)

`d)`

\(\left(-\dfrac{2}{3}x-\dfrac{3}{2}\right)\left(\dfrac{3}{-2}-\dfrac{10}{3}\right)=\dfrac{29}{5}\\ \Rightarrow\left(-\dfrac{2}{3}x-\dfrac{3}{2}\right)\cdot\left(-\dfrac{29}{6}\right)=\dfrac{29}{5}\\ \Rightarrow-\dfrac{2}{3}x-\dfrac{3}{2}=\dfrac{29}{5}\div\left(-\dfrac{29}{6}\right)\\ \Rightarrow-\dfrac{2}{3}x-\dfrac{3}{2}=\dfrac{29}{5}\cdot\left(-\dfrac{6}{29}\right)\\ \Rightarrow-\dfrac{2}{3}x-\dfrac{3}{2}=-\dfrac{6}{5}\\ \Rightarrow-\dfrac{2}{3}x=-\dfrac{6}{5}+\dfrac{3}{2}\\ \Rightarrow-\dfrac{2}{3}x=\dfrac{3}{10}\\ \Rightarrow x=\dfrac{3}{10}\div\left(-\dfrac{2}{3}\right)\\ \Rightarrow x=-\dfrac{9}{20}\)

Vậy, \(x=-\dfrac{9}{20}\)

`15.`

Tỉ số giữa chiều cao nhà của Hùng và cây bàng trước nhà:

\(\dfrac{49}{4}\div\dfrac{17}{4}\\ =\dfrac{49}{4}\cdot\dfrac{4}{17}\\ =\dfrac{49}{17}\)

Vậy, tỉ số chiều cao giữa nhà của Hùng và cây bàng trước nhà là \(\dfrac{49}{17}\)

`16.`

\(A=\dfrac{3x-2}{x+3}\\ =\dfrac{3x+9-11}{x+3}\\ =\dfrac{3x+9}{x+3}-\dfrac{11}{x+3}\\ =\dfrac{3\left(x+3\right)}{x+3}-\dfrac{11}{x+3}\\ =3-\dfrac{11}{x+3}\)

Để A nguyên:

\(3-\dfrac{11}{x+3}\in Z\\ \Rightarrow\dfrac{11}{x+3}\in Z\)

$\Rightarrow $ `11 \vdots (x + 3)`

$\Rightarrow $ `x + 3 \in \text{Ư(11)} = {+-1; +-11}`

\(\Rightarrow\left[{}\begin{matrix}x+3=1\\x+3=-1\\x+3=11\\x+3=-11\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-2\\x=-4\\x=8\\x=-14\end{matrix}\right.\)

Vậy, để A nguyên, `x \in {-2; -4; 8; -14}.`

`#3107.101107`

`12.`

`a)`

\(\left(\dfrac{3}{7}\right)^{21}\div\left(\dfrac{9}{49}\right)^6\\ =\left(\dfrac{3}{7}\right)^{21}\div\left(\dfrac{3^2}{7^2}\right)^6\\= \left(\dfrac{3}{7}\right)^{21}\div\left(\dfrac{3}{7}\right)^{2\cdot6}\\ =\left(\dfrac{3}{7}\right)^{21-2\cdot6}\\ =\left(\dfrac{3}{7}\right)^{21-12}\\ \left(\dfrac{3}{7}\right)^{19}\)

`b)`

\(3-\left(-\dfrac{6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2\div2\\ =3-1+\left(\dfrac{1}{2}\right)^2\cdot\dfrac{1}{2}\\ =2+\left(\dfrac{1}{2}\right)^3\\ =2+\dfrac{1}{8}\\ =\dfrac{17}{8}\)

`13.`

Gọi x là số trang sách của quyển sách

Số trang sách Hùng đọc ngày thứ nhất: \(\dfrac{1}{3}\cdot x\)

Số trang sách Hùng đọc ngày thứ `2`:\(\dfrac{5}{8}\cdot\left(x-\dfrac{1}{3}x\right)=\dfrac{5}{8}\cdot\dfrac{2}{3}x=\dfrac{5}{12}x\) 

Số trang sách ngày thứ `3` Hùng đọc là `90` trang

Ta có:

\(x-\dfrac{1}{3}x-\dfrac{5}{12}x=90\)

\(\Rightarrow x\left(1-\dfrac{1}{6}-\dfrac{5}{12}\right)=90\\ \Rightarrow\dfrac{5}{12}x=90\\ \Rightarrow x=90\div\dfrac{5}{12}\\ \Rightarrow x=90\cdot\dfrac{12}{5}\\ \Rightarrow x=216\)

Vậy, số trang của quyển sách đó là `216.`

`#3107.101107`

`1.`

`a)`

\(\dfrac{x}{10}=\dfrac{y}{6}=\dfrac{z}{21}\); \(6x-2z=28\)

\(\Rightarrow\dfrac{6x}{60}=\dfrac{y}{6}=\dfrac{2z}{42}\)

AD tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{6x}{60}=\dfrac{y}{6}=\dfrac{2z}{42}=\dfrac{6x-2z}{60-42}=\dfrac{28}{18}=\dfrac{14}{9}\)

\(\Rightarrow\dfrac{x}{10}=\dfrac{y}{6}=\dfrac{z}{21}\\ \Rightarrow\left\{{}\begin{matrix}x=10\cdot\dfrac{14}{9}=\dfrac{140}{9}\\y=6\cdot\dfrac{14}{9}=\dfrac{28}{3}\\z=21\cdot\dfrac{14}{9}=\dfrac{98}{3}\end{matrix}\right.\)

Vậy, \(x=\dfrac{140}{9};y=\dfrac{28}{3};z=\dfrac{98}{3}\)

`b)`

\(3x=2y;7y=5z;x-y+z=32?\)

Ta có:

\(\dfrac{x}{2}=\dfrac{y}{3};\dfrac{y}{5}=\dfrac{z}{7}\\ \Rightarrow\dfrac{x}{2}=\dfrac{5y}{15};\dfrac{3y}{15}=\dfrac{z}{7}\\ \Rightarrow\dfrac{x}{10}=\dfrac{y}{15};\dfrac{y}{15}=\dfrac{z}{21}\\ \Rightarrow\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{21}\)

AD tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{21}=\dfrac{x-y+z}{10-15+21}=\dfrac{32}{16}=2\)

\(\Rightarrow\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{21}=2\\ \Rightarrow\left\{{}\begin{matrix}x=10\cdot2=20\\y=15\cdot2=30\\z=21\cdot2=42\end{matrix}\right.\)

Vậy, `x = 20; y = 30; z = 42`

`c)`

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