\(b,\dfrac{2}{3}\left(2x+4\right)^2-\dfrac{1}{3}\left(x+1\right)^2=-\dfrac{1}{3}\left(2x+4\right)^2+\dfrac{2}{3}\left(x+1\right)^2\)

\(\Rightarrow\dfrac{2}{3}\left(2x+4\right)^2+\dfrac{1}{3}\left(2x+4\right)^2=\dfrac{2}{3}\left(x+1\right)^2+\dfrac{1}{3}\left(x+1\right)^2\)

\(\Rightarrow\left(2x+4\right)^2=\left(x+1\right)^2\)

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

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

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

\(\Rightarrow\left[{}\begin{matrix}x=-3\\3x=-5\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{5}{3}\end{matrix}\right.\)

Vậy: \(x\in\left\{-3;-\dfrac{5}{3}\right\}\).

\(\left[\left(\dfrac{-3}{8}+\dfrac{11}{23}\right):\dfrac{5}{9}+\left(\dfrac{-5}{8}+\dfrac{12}{23}\right):\dfrac{5}{9}\right]\cdot\dfrac{11}{235}\)

\(=\left[\left(\dfrac{-3}{8}+\dfrac{11}{23}\right)\cdot\dfrac{9}{5}+\left(\dfrac{-5}{8}+\dfrac{12}{23}\right)\cdot\dfrac{9}{5}\right]\cdot\dfrac{11}{235}\)

\(=\dfrac{9}{5}\cdot\left(\dfrac{-3}{8}+\dfrac{11}{23}+\dfrac{-5}{8}+\dfrac{12}{23}\right)\cdot\dfrac{11}{235}\)

\(=\dfrac{9}{5}\cdot\left[\left(\dfrac{-3}{8}+\dfrac{-5}{8}\right)+\left(\dfrac{11}{23}+\dfrac{12}{23}\right)\right]\cdot\dfrac{11}{235}\)

\(=\dfrac{9}{5}\cdot\left(-1+1\right)\cdot\dfrac{11}{235}\)

\(=\dfrac{9}{5}\cdot0\cdot\dfrac{11}{235}\)

\(=0\)

In spite of loving ice-cream, Linda didn't eat more than one

Mọi người hỏi các bài toán của lớp 4 đi

One of my favorite hobby is playing football. I spend a lot of free time playing football after finishing my homework. I was so interested in playing football from my childhood and started learning to play when I was 9 years old. When I was 10 years old, my dad told my teacher about my hobby of football. My teacher told my dad that there was a facility for playing sports daily in school so he could admit his child. Now, I enjoy playing football and I participate in inter-school competitions.

a) 2/7 : x = 11/6 : 7/12

2/7 : x = 22/7

x = 2/7 : 22/7

x = 1/11

b) (2 - x)/3 = -3/(x - 2)

(2 - x)(x - 2) = -3.3

-(x - 2)² = -9

(x - 2)² = 9

x - 2 = 3 hoặc x - 2 = -3

*) x - 2 = 3

x = 3 + 2

x = 5

*) x - 2 = -3

x = -3 + 2

x = -1

Vậy x = -1; x = 5

c) (x - 1)/(x + 2) = 2/3

3(x - 1) = 2(x + 2)

3x - 3 = 2x + 4

3x - 2x = 4 + 3

x = 7

a) 2/7 : x = 11/6 : 7/12

2/7 : x = 22/7

x = 2/7 : 22/7

x = 1/11

b) (2 - x)/3 = -3/(x - 2)

(2 - x)(x - 2) = -3.3

-(x - 2)² = -9

(x - 2)² = 9

x - 2 = 3 hoặc x - 2 = -3

*) x - 2 = 3

x = 3 + 2

x = 5

*) x - 2 = -3

x = -3 + 2

x = -1

Vậy x = -1; x = 5

c) (x - 1)/(x + 2) = 2/3

3(x - 1) = 2(x + 2)

3x - 3 = 2x + 4

3x - 2x = 4 + 3

x = 7

Lời giải:

Gọi số học sinh là $a$ và số hàng là $b$. Theo bài ra ta có:

$ab=800$

$\frac{a}{10}=\frac{b}{8}$

$\Rightarrow (\frac{a}{10})^2=(\frac{b}{8})^2=\frac{a.b}{10.8}=\frac{800}{80}=10$

$\Rightarrow \frac{a}{10}=\sqrt{10}$ (vô lý)

Đề có vấn đề. Bạn xem lại.

a) x² - 2 = 0

x² = 2

x = -√2 (loại) hoặc x = √2 (loại)

Vậy không tìm được x Q thỏa mãn đề bài

b) x² + 7/4 = 23/4

x² = 23/4 - 7/4

x² = 4

x = 2 (nhận) hoặc x = -2 (nhận)

Vậy x = -2; x = 2

c) (x - 1)² = 0

x - 1 = 0

x = 1 (nhận)

Vậy x = 1

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