1-2-3-4-...-1 0000...0 hơi sai sai. Số bé làm sao trừ được số lớn, lớp 5 cũng chưa học số âm?

1) \(2x^4+3x^3-x^2+3x+2=0\)

\(\Rightarrow2x^4+x^3+2x^3+x^2-2x^2-x+4x+2=0\)

\(\Rightarrow x^3\left(2x+1\right)+x^2\left(2x+1\right)-x\left(2x+1\right)+2\left(2x+1\right)=0\)

\(\Rightarrow\left(2x+1\right)\left(x^3+x^2-x+2\right)=0\)

\(\Rightarrow\left(2x+1\right)\left(x^3+2x^2-x^2-2x+x+2\right)=0\)

\(\Rightarrow\left(2x+1\right)\left[x^2\left(x+2\right)-x\left(x+2\right)+\left(x+2\right)\right]=0\)

\(\Rightarrow\left(2x+1\right)\left(x+2\right)\left(x^2-x+1\right)=0\)

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

Ta có:

\(x^2-x+1\)

\(=x^2-2x.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}+1\)

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

Vì \(\left(x-\dfrac{1}{2}\right)^2\ge0\) với mọi x

\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\) với mọi x

\(\Rightarrow x^2-x+1\) vô nghiệm

\(\Rightarrow\left[{}\begin{matrix}2x+1=0\\x+2=0\end{matrix}\right.\)

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

3) \(\left(x+2\right)^4+\left(x+4\right)^4=16\)

Đặt x + 3 = a, ta được

\(\left(a-1\right)^4+\left(a+1\right)^4=16\)

\(\Rightarrow\left[\left(a-1\right)^2\right]^2+\left[\left(a+1\right)^2\right]^2=16\)

\(\Rightarrow\left(a^2-2a+1\right)^2+\left(a^2+2a+1\right)^2=16\)

\(\Rightarrow a^4+4a^2+1+2a^2-4a^3-4a+a^4+4a^2+1+2a^2+4a^3+4a=16\)

\(\Rightarrow2a^4+2.4a^2+2+2.2a^2=16\)

\(\Rightarrow2a^4+8a^2+4a^2+2=16\)

\(\Rightarrow2a^4+12a^2+2-16=0\)

\(\Rightarrow2a^4+12a^2-14=0\)

\(\Rightarrow2a^4-2a^2+14a^2-14=0\)

\(\Rightarrow2a^2\left(a^2-1\right)+14\left(a^2-1\right)=0\)

\(\Rightarrow\left(a^2-1\right)\left(2a^2+14\right)=0\)

\(\Rightarrow\left(a-1\right)\left(a+1\right).2\left(a^2+7\right)=0\)

\(\Rightarrow\left(a-1\right)\left(a+1\right)\left(a^2+7\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}a-1=0\\a+1=0\\a^2+7=0\end{matrix}\right.\)

Vì \(a^2\ge0\) với mọi a

\(\Rightarrow a^2+7\ge7\) với mọi a

\(\Rightarrow a^2+7\) vô nghiệm

\(\Rightarrow\left[{}\begin{matrix}a-1=0\\a+1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x+3-1=0\\x+3+1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x+2=0\\x+4=0\end{matrix}\right.\)

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

a)(Sửa đề) \(4(x^2-6x+9)-16(4x^2+28x+49)=0\)

\(⇔(2x-6)^2-(8x+28)^2=0\)

\(⇔(-6x-34)(10x+22)=0\)

\(⇔\left[\begin{array}{} -6x-34=0\\ 10x+22=0 \end{array}\right.\)

\(⇔\left[\begin{array}{} x=-\dfrac{17}{3}\\ x=-\dfrac{11}{5} \end{array}\right.\)

b)(Sửa đề 1) \((2x-16)^2-(x-4)^2=0\)

\(⇔(3x-20)(x-12)=0\)

\(⇔\left[\begin{array}{} 3x-20=0\\ x-12=0 \end{array}\right.\)\(⇔\left[\begin{array}{} x=\frac{20}{3}\\ x=12 \end{array}\right.\)

(Sửa đề 2) \((x^2-16)^2-(x-4)^2=0\)

\(⇔(x^2-x-12)(x^2+x-20)=0\)

\(⇔(x-4)^2(x+3)(x+5)=0\)

\(⇔\left[\begin{array}{} (x-4)^2=0\\\ x+3=0\\ x+5=0 \end{array}\right.\)\(⇔\left[\begin{array}{} x=4\\\ x=-3\\ x=-5 \end{array}\right.\)

A=\(2^2-9^3+4^{-2}.16-2.5^2\)
\(=4-729+1-50=-774\)
B=\(\left(2^3.2\right).\dfrac{1}{2}+3^{-2}.3^2-7.1+5\)
\(B=2^4.\dfrac{1}{2}+1-7+5=8+1-7+5=7\)
 

 C = 2^-3 + (5²)³.5^-3 + 4^-3.16 - 2.3² - 105.(\(\dfrac{24}{51}\))⁰

C =  \(\dfrac{1}{8}\) + 5⁶.5^-3 + 4^-3.4² - 2.9 - 105.1

C =  \(\dfrac{1}{8}\) + 5³ + \(\dfrac{1}{4}\) - 18 - 105

C =  (\(\dfrac{1}{8}\) + \(\dfrac{1}{4}\))  - (105 - 125 + 18)

C = \(\dfrac{3}{8}\) - (-20 + 18)

C = \(\dfrac{3}{8}\)  + 2

C = \(\dfrac{19}{8}\)

D = 2^-3 . \(\dfrac{1}{2^{-2}}\).\(\dfrac{2}{3}\) + 4^-2.8 - 7. (\(\dfrac{17}{23}\))⁰ + 19

D = 2^-1.\(\dfrac{2}{3}\) + \(\dfrac{1}{16}\).8 - 7.1 + 19

D = \(\dfrac{1}{3}\) + \(\dfrac{1}{2}\) - 7 + 19

D = \(\dfrac{5}{6}\)  + (19 - 7)

D = \(\dfrac{5}{6}\) + 12

D = \(\dfrac{77}{6}\)

a) \(\left(x-4\right)\left(x+3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-4=0\\x+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=4\\x=-3\end{cases}}\)

Vậy \(x\in\left\{-3;4\right\}\)

b)\(\left(x^2+16\right)\left(x^2-16\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+16=0\\x^2-16=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\sqrt{-16}\\x=\sqrt{16}=4\end{cases}}\)

Vậy \(x=4\)

\(\left(x-4\right)\left(x+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x+3=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=4\\x=-3\end{cases}}\)

\(\left(x^2+16\right)\left(x^2-16\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+16=0\\x^2-16=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x^2=-16\left(loại\right)\\x^2=16\end{cases}}\Rightarrow x=\left(\pm4\right)^2\)

\(\left(x-2\right)\left(x+4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+4=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-4\end{cases}}\)

c) \(\left(x^2+10\right)\left(x-3\right)< 0\)

\(\Rightarrow\hept{\begin{cases}x^2+10\\x-3\end{cases}}\)trái dấu

TH1 : \(\hept{\begin{cases}x^2+10>0\\x-3< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}x^2>-10\\x< 3\end{cases}}\Leftrightarrow\hept{\begin{cases}x\inℤ\\x< 3\end{cases}}\Leftrightarrow x\in\left\{...;1;2\right\}\)

TH2 : \(\hept{\begin{cases}x^2+10< 0\\x-3>0\end{cases}}\Leftrightarrow\hept{\begin{cases}x^2< -10\\x>3\end{cases}}\Leftrightarrow\hept{\begin{cases}x\in\varnothing\\x>3\end{cases}}\Leftrightarrow x\in\left\{\varnothing\right\}\)

\(\Rightarrow x\in\left\{...;1;2\right\}\)