Giải:

\(x\left(x+1\right)+\left(x+2\right)+\left(x+3\right)=24\)

\(\Leftrightarrow x^2+x+x+2+x+3=24\)

\(\Leftrightarrow x^2+3x+5=24\)

\(\Leftrightarrow x^2+3x=19\)

\(\Leftrightarrow x\left(x+3\right)=19\)

Ta có bảng:

Vậy ...

cả 2 pt đều giải theo kiểu cái đầu nhóm với cái cuối, 2 cái ở giữa nhóm với nhau. sau đó giải theo cách đặt ẩn phụ

1) \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=0\)

\(\Leftrightarrow\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24=0\)

\(\Leftrightarrow\left(x^2+5x+2x+10\right)\left(x^2+4x+3x+12\right)-24=0\)

\(\Leftrightarrow\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24=0\)

Đặt \(x^2+7x=a\), nên ta có :

\(\left(a+10\right)\left(a+12\right)-24=0\)

\(\Leftrightarrow\left(x+11-1\right)\left(x+11+1\right)-24=0\)

\(\Leftrightarrow\left[\left(x+11\right)^2-1\right]-24=0\)

\(\Leftrightarrow\left(x+11\right)^2-25=0\)

\(\Leftrightarrow\left(x+11-5\right)\left(x+11+5\right)=0\)

\(\Leftrightarrow\left(x+6\right)\left(x+16\right)=0\Leftrightarrow\orbr{\begin{cases}x=-6\\x=-16\end{cases}}\)

1,(x+2)(x+5)(x+3)(x+4)-24=(x²+7x+10)(x²+7x+12)-24

Đặt x²+7x+10= t ta có t(t+2)-24=t²+2t-24=(t-4)(t+6)

hay (x²+7x+6)(x²+7x+16)

2,x(x+10)(x+4)(x+6)+128=(x²+10x)(x²+10x+24)+128

Đặt x²+10x=t ta có t(t+24)+128=t²+24t+128=(t+8)(t+16)

hay (x²+10x+8)(x²+10x+16)

3,(x+2)(x-7)(x+3)(x-8)-144=(x^2-5x-14)(x²-5x-24)-144

Đặt x²-5x-14=t ta có t(t-10)-144=t²-10t-144=(t-18)(t+8)

Hay (x²-5x-32)(x²-5x-6)=(x²-5x-32)(x+1)(x-6)

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a)\((x^2- 4).(x^2 - 10) = 72 Đặt x^2 - 7 = a(1), ta có (a+3)(a-3)=72 a^2-9=72 a^2=81 a=+-9 xét 2 trường hợp a = 9 và -9 khi thay vào (1) ta có..... tự lm nốt nha \)

b) nhóm x+1 vs x+4 và x+2 vs x+3 ta sẽ có (x²+5x+4)(x²+5x+6)(x+5)=40

câu c dễ lắm... bạn có thể làm 2 cách 1 là xét 2 th khi vế trái bằng vế phải hay trái dấu vế phải

hoặc cách 2 đưa về hiệu 2 bình phương nhá.. cách này dễ hơn

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

⇔\(3x^2\)-6x-\(4x^2\)+2x=1-\(x^2\)

⇔-1\(x^2\) - 4x= 1- \(x^2\)

⇔ -1\(x^2\) -4x+ \(x^2\) = 1

⇔-4x=1

⇔ x = \(\dfrac{-1}{4}\)

1.\(\left(x-5\right).\left(x+5\right)-\left(x+3\right)^2=2x-3\)

\(\Leftrightarrow x^2-25-\left(x^2+6x+9\right)=2x-3\)

\(\Leftrightarrow x^2-25-x^2-6x-9=2x-3\)

\(\Leftrightarrow x^2-25-x^2-6x-9-2x+3=0\)

\(\Leftrightarrow-8x-31=0\)

\(\Leftrightarrow x=\dfrac{-31}{8}\)

\(\left(x-4\right)^3-\left(x-5\right)\left(x^2+5x+25\right)=\left(x+2\right)\left(x^2-2x+4\right)-\left(x+4\right)^3\)

\(\Leftrightarrow\left(x-4\right)^3-\left(x^3-5^3\right)=\left(x^3+2^3\right)-\left(x+4\right)^3\)

\(\Leftrightarrow\left(x-4\right)^3-x^3+5^3=x^3+2^3-\left(x+4\right)^3\)

\(\Leftrightarrow\left(x^3-12x^2+48x-64\right)-x^3+5^3=x^3+2^3-\left(x^3+12x^2+48x+64\right)\)

\(\Leftrightarrow x^3-12x^2+48x-64-x^3+5^3=x^3+2^3-x^3-12x^2-48x-64\)

\(\Leftrightarrow-12x^2+48x-64+5^3=2^3-12x^2-48x-64\)

\(\Leftrightarrow-12x^2+48x-61=-12x^2-48x-56\)

\(\Leftrightarrow96x=-117\)

\(\Leftrightarrow x=\dfrac{-117}{96}=\dfrac{-39}{32}\)

2. \(\left(2x+3\right)^2+\left(x-1\right)\left(x+1\right)=5\left(x+2\right)^2\)

\(\Leftrightarrow4x^2+12x+9+x^2-1=5\left(x^2+4x+4\right)\)

\(\Leftrightarrow4x^2+12x+9+x^2-1=5x^2+20x+20\)

\(\Leftrightarrow4x^2+x^2-5x^2+12x-20x=20-9+1\)

\(\Leftrightarrow-8x=12\)

\(\Leftrightarrow x=\dfrac{-12}{8}=\dfrac{-3}{2}\)

1. a, 3x + |x - 2| = 8
<=> |x - 2| = 8 - 3x
Xét 2 TH :
TH1: x - 2 = 8 - 3x
<=> x + 3x = 8 + 2
<=> 4x = 10
<=> x = \(\dfrac{5}{2}\) (thỏa mãn)
TH2: x - 2 = -(8 - 3x)
<=> x - 2 = -8 + 3x
<=> -2 + 8 = 3x - x
<=> 6 = 2x
<=> x = 3 (thỏa mãn)
b, 5 - |x - 1| = 4
<=> |x - 1| = 1
<=> \(\left[{}\begin{matrix}x-1=1\\x-1=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\) (thỏa mãn)
@Nguyễn Hoàng Vũ

2. 5.(x - 2) - 4.(1 - 3x) = |3 - 7| + 2.(1 + 2x)
<=> 5x - 10 - 4 + 12x = 4 + 2 + 4x
<=> 17x - 14 = 6 + 4x
<=> 17x - 4x = 6 + 14
<=> 13x = 20
<=> x = \(\dfrac{20}{13}\) (thỏa mãn)
@Nguyễn Hoàng Vũ

4. 1\(\dfrac{1}{2}\).(x - 2) - \(\dfrac{x-5}{3}\) = 3\(\dfrac{1}{3}\).(1 - 2x) - \(\dfrac{5.\left(x+1\right)}{6}\)

<=> \(\dfrac{3}{2}\).(x - 2) - \(\dfrac{x-5}{3}\) = \(\dfrac{10}{3}\).(1 - 2x) - \(\dfrac{5x+5}{6}\)

<=> \(\dfrac{3}{2}x-3-\dfrac{x}{3}+\dfrac{5}{3}=\dfrac{10}{3}-\dfrac{20}{3}x-\dfrac{5x}{6}-\dfrac{5}{6}\)

<=> \(\dfrac{3}{2}x-\dfrac{x}{3}+\dfrac{20}{3}x-\dfrac{5x}{6}=\dfrac{10}{3}-\dfrac{5}{6}-3+\dfrac{5}{3}\)

<=> 7x = \(\dfrac{7}{6}\)

<=> x = \(\dfrac{1}{6}\)
@Nguyễn Hoàng Vũ

\(\left(x-3\right).\left(y-2\right)=7\)

\(\Rightarrow x-3;y-2\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\)

 Ta có bảng sau:

Vậy (x;y) là :(-4;1),(2;5),(4;7),(10;3)

làm tương tự