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\(a.\left(x^2+3x+2\right)\left(x^2+11x+30\right)-60=0\)
\(\Leftrightarrow\left(x^2+7x-4x+16-14\right)\left(x^2+7x+4x+16+14\right)-60=0\)
\(\Leftrightarrow\left(x^2+7x+16-4x-14\right)\left(x^2+7x+16+4x+14\right)=0\)
\(\Leftrightarrow\left(x^2+7x+16\right)^2-\left(4x+14\right)^2-60=0\)
Vì \(\left(x^2+7x+16\right)^2>0;\left(4x+14\right)^2>0\)
Nên \(\left(x^2+7x+16\right)^2-\left(4x+14\right)^2-60\ge-60\)
V...\(S=\varnothing\)
\(b.4^x-12.2^x+32=0\)
\(\Leftrightarrow\left(2^x\right)^2-2.2^x.6+36-4=0\)
\(\Leftrightarrow\left(2^x-6\right)^2-4=0\)
\(\Leftrightarrow\left(2^x-4\right)\left(2^x-8\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2^x-4=0\\2^x-8=0\end{cases}\Leftrightarrow\orbr{\begin{cases}2^x=4\\2^x=8\end{cases}\Leftrightarrow}\orbr{\begin{cases}2^x=2^2\\2^x=2^3\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=3\end{cases}}}\)
V...\(S=\left\{2;3\right\}\)
^^ đúng ko ta
a) (x+1)(x+2)(x+5)(x+6)-60=0
[(x+1)(x+6)][(x+2)(x+5)]-60=0
(x^2 + 7x + 6)(x^2 + 7x + 10) - 60 = 0
đặt t = x^2 + 7x + 8
pt trở thành
(t-2)(t+2)-60=0
t^2 - 64=0 .....
t=8 hoặc t=-8.
tìm x ....
\(ĐKXĐ:x\ne-4;x\ne-5;x\ne-6;x\ne-7\)
\(\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}\)
\(\Rightarrow\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)
\(\Rightarrow\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{18}\)
\(\Rightarrow\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\)
\(\Rightarrow\frac{3}{\left(x+4\right)\left(x+7\right)}=\frac{3}{54}\)
\(\Rightarrow\left(x+4\right)\left(x+7\right)=54\)
\(\Leftrightarrow x^2+11x+28=54\)
\(\Leftrightarrow x^2+11x-26=0\)
Ta có \(\Delta=11^2+4.26=225,\sqrt{\Delta}=15\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{-11+15}{2}=2\\x=\frac{-11-15}{2}=-13\end{cases}}\)
Vậy tập nghiệm S = {2;-13}
a, \(x^4-6x^3+11x^2-6x+1=0\)
\(\Rightarrow\left(x^2-3x+1\right)^2=0\)
\(\Rightarrow x^2-3x+1=0\)
\(\Rightarrow x=\frac{\pm\sqrt{5}+3}{2}\)
Chúc bạn học tốt
\(x^4-\left(6x^2-2x^2\right)+\left(9x^2-6x+1\right)=0\)
\(x^4-2x^2\left(3x-1\right)+\left(3x-1\right)^2=0\)
\(\left(x^2-3x+1\right)^2=0\)
tự làm
B) \(\left(6x^4-18x^3\right)+\left(13x^{^3}-39x^2\right)+\left(x-3x\right)-\left(2x-6\right)=0\)
\(6x^3\left(x-3\right)+13x^2\left(x-3\right)+x\left(x-3\right)-2\left(x-3\right)=0\)
\(\left(x-3\right)\left(6x^3+13x^2-2\right)=0\)
\(\left(x-3\right)\left(6x^3+12x^2+x^2+2x-x-2\right)\)
\(\left(x-3\right)\left\{6x^2\left(x+2\right)+x\left(x+2\right)-\left(x+2\right)\right\}\)
\(\left(x-3\right)\left(x+2\right)\left(6x^2-x-1\right)\)
\(\left(x-3\right)\left(x+2\right)\left(6x^2-3x+2x-1\right)\)
\(\left(x-3\right)\left(x+2\right)\left(3x\left(2x-1\right)+\left(2x-1\right)\right)\)
\(\left(x-3\right)\left(x+2\right)\left(2x-1\right)\left(3x+1\right)=0\)
câu C nghĩ đã
a)\(\left(x^2+1\right)\left(x^2-4x+4\right)=0\Leftrightarrow\orbr{\begin{cases}x^2+1=0\\x^2-4x+4=0\end{cases}\Rightarrow\orbr{\begin{cases}x^2=-1\left(vn\right)\\\left(x-2\right)^2=0\end{cases}\Rightarrow}x=2}\)
b)\(\left(3x-2\right)\left(\frac{2x+6}{7}-\frac{4x-3}{5}\right)=0\\ \Rightarrow\left(3x-2\right)\left(\frac{10x+30-28x+21}{35}\right)=0\)
\(\Rightarrow\left(3x-2\right)\left(\frac{-18x+51}{35}\right)=0\Rightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{17}{6}\end{cases}}\)
c)\(\left(3,3-11x\right)\left(\frac{21x+6+10-30x}{15}\right)=0\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{10}\\x=\frac{16}{9}\end{cases}}\)
1b)
Đặt \(\overline{abcd}=k^2\left(k\in N;32\le k\le99\right)\)
Note : nếu k nằm ngoài khoảng giá trị ở trên thì k2 sẽ có ít hơn hoặc nhiều hơn 4 chữ số
Theo bài cho :
\(\overline{ab}-\overline{cd}=1\Rightarrow\overline{ab}=\overline{cd}+1\Rightarrow\overline{abcd}=k^2\Leftrightarrow100\cdot\overline{ab}+\overline{cd}=k^2\)
\(\Leftrightarrow100\cdot\overline{cd}+100+\overline{cd}=k^2\Leftrightarrow101\cdot\overline{cd}=k^2-100\Leftrightarrow101\overline{cd}=\left(k-10\right)\left(k+10\right)\)
\(\Rightarrow\orbr{\begin{cases}k-10⋮101\\k+10⋮101\end{cases}}\)
Mà \(\text{ }(k-10;101)=1\Rightarrow k+10⋮101\)
Lại có : \(32\le k\le99\Rightarrow42\le k+10\le109\)
\(\Rightarrow k+10=101\Rightarrow k=91\Rightarrow\overline{abcd}=91^2=8182\left(tm\right)\)
a) 7x - 35 = 0
<=> 7x = 0 + 35
<=> 7x = 35
<=> x = 5
b) 4x - x - 18 = 0
<=> 3x - 18 = 0
<=> 3x = 0 + 18
<=> 3x = 18
<=> x = 5
c) x - 6 = 8 - x
<=> x - 6 + x = 8
<=> 2x - 6 = 8
<=> 2x = 8 + 6
<=> 2x = 14
<=> x = 7
d) 48 - 5x = 39 - 2x
<=> 48 - 5x + 2x = 39
<=> 48 - 3x = 39
<=> -3x = 39 - 48
<=> -3x = -9
<=> x = 3
1/ Ta có
\(x^2+9x+20=x^2+4x+5x+20=x\left(x+4\right)+5\left(x+4\right)=\left(x+4\right)\left(x+5\right)\)
Tương tự
\(x^2+11x+30=\left(x+5\right)\left(x+6\right)\)
\(x^2+13x+42=\left(x+6\right)\left(x+7\right)\)
Đk: x khác 4, 5, 6, 7
\(\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)
\(\Leftrightarrow\frac{\left(x+5\right)-\left(x+4\right)}{\left(x+4\right)\left(x+5\right)}+\frac{\left(x+6\right)-\left(x+5\right)}{\left(x+5\right)\left(x+6\right)}+\frac{\left(x+7\right)-\left(x+6\right)}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)
\(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{18}\)
\(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\) EM tự làm tiếp nhé
x 2 + (x + 2)(11x – 7) = 4
⇔ x 2 – 4 + (x + 2)(11x – 7) = 0
⇔ (x + 2)(x – 2) + (x + 2)(11x – 7) = 0
⇔ (x + 2)[(x – 2) + (11x – 7)] = 0
⇔ (x + 2)(x – 2 + 11x – 7) = 0
⇔ (x + 2)(12x – 9) = 0 ⇔ x + 2 = 0 hoặc 12x – 9 = 0
x + 2 = 0 ⇔ x = - 2
12x – 9 = 0 ⇔ x = 0,75
Vậy phương trình có hai nghiệm x = - 2 hoặc x = 0,75