Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(\left(2x+3\right)^2-3\left(x-4\right)\left(x+4\right)=\left(x-2\right)^2+1\)
\(\Leftrightarrow4x^2+12x+9-3\left(x^2-16\right)=x^2-4x+4+1\)
\(\Leftrightarrow4x^2+12x+9-3x^2+48=x^2-4x+5\)
\(\Leftrightarrow x^2+12x+57=x^2-4x+5\)
\(\Leftrightarrow16x+52=0\)
\(\Leftrightarrow x=-\frac{13}{4}\)
b) \(\left(3x-2\right)\left(9x^2+6x+4\right)-\left(3x-1\right)\left(9x^2-3x+1\right)=x-4\)
\(\Leftrightarrow\)Xem lại đề !
c) \(x\left(x-1\right)-\left(x-3\right)\left(x+4\right)=5x\)
\(\Leftrightarrow x^2-x-x^2-x+12=5x\)
\(\Leftrightarrow-2x+12=5x\)
\(\Leftrightarrow7x-12=0\)
\(\Leftrightarrow x=\frac{12}{7}\)
d) \(\left(2x+1\right)\left(2x-1\right)=4x\left(x-7\right)-3x\)
\(\Leftrightarrow4x^2-1=4x^2-28x-3x\)
\(\Leftrightarrow28x+3x-1=0\)
\(\Leftrightarrow31x-1=0\)
\(\Leftrightarrow x=\frac{1}{31}\)
1) \(2x.\left(x-7\right)-\left(x+3\right)\left(x-2\right)-\left(x+4\right)\left(x-4\right)\)
\(=2x^2-14x-\left(x^2+x-6\right)-\left(x^2-4\right)\)
\(=-15x+10\)
b) \(2x.\left(x+1\right)^2-\left(x-1\right)^3-\left(x-2\right)\left(x^2+2x+4\right)\)
\(=2x.\left(x^2+2x+1\right)-\left(x^3-3x^2+3x-1\right)-\left(x^3-8\right)\)
\(=2x^3+4x^2+2x-x^3+3x^2-3x+1-x^3+8\)
\(=7x^2-x+9\)
c) \(\left(x-5\right)\left(x+5\right)\left(x+2\right)-\left(x+2\right)^3\)
\(=\left(x+2\right).\left[\left(x-5\right)\left(x+5\right)-\left(x+2\right)^2\right]\)
\(=\left(x+2\right).\left(x^2-25-x^2-4x-4\right)\)
\(=\left(x+2\right)\left(-4x-29\right)\)
\(=-4x^2-37x-58\)
d) \(\left(x-3\right)^3+\left(x-5\right)\left(x^2+5x+25\right)-\left(x-1\right)\left(x^2+x+1\right)\)
\(=x^3-9x^2+27x-27+\left(x^3-125\right)-\left(x^3-1\right)\)
\(=x^3-9x^2+27x-151\)
e) \(\left(x-1\right)^3-\left(x-2\right)\left(x^2-2x+4\right)+3x^2+2x\)
\(=x^3-3x^2+3x-1-\left(x^3-8\right)+3x^2+2x\)
\(=5x+7\)
Nhẩm ấy, ko nháp âu
\(2x\left(x-7\right)-\left(x+3\right)\left(x-2\right)-\left(x+4\right)\left(x-4\right)\)
\(=2x^2-14x-\left(x^2-2x+3x-6\right)-\left(x^2-4x+4x-16\right)\)
\(=2x^2-14x-x^2+x-6-x^2+16\)
\(=-13x-10\)
\(2x\left(x+1\right)^2-\left(x-1\right)^3-\left(x-2\right)\left(x^2+2x+4\right)\)
\(=2x\left(x^2+2x+1\right)-\left(x^3-3x^2+3x-1\right)-\left(x-2\right)\left(x+2\right)\)
\(-2x^3+4x^2+2x-x^3+3x^2-3x+1-x^2+4\)
\(=-3x^3+6x^2-x+5\)
1. (x + 2)(x2 - 2x + 4) - (x3 + 2x2) = 5
=> x(x2 - 2x + 4) + 2(x2 - 2x + 4) - x3 - 2x2 - 5 = 0
=> x3 - 2x2 + 4x + 2x2 - 4x + 8 - x3 - 2x2 - 5 = 0
=> (x3 - x3) + (-2x2 + 2x2 - 2x2) + (4x - 4x) + (8 - 5) = 0
=> -2x2 + 3 = 0
=> -2x2 = -3
=> x2 = 3/2
=> x = \(\pm\sqrt{\frac{3}{2}}\)
2. \(\left(x+5\right)^2-6=0\)
=> x2 + 10x + 25 - 6 = 0
=> x2 + 10x + 19 = 0
=> x vô nghiệm(do mình không để căn nên ghi vô nghiệm thôi nhá)
3. \(\left(x+3\right)\left(x^2-3x+9\right)-x^3=2x\)
=> x(x2 - 3x + 9) + 3(x2 - 3x + 9) - x3 - 2x = 0
=> x3 - 3x2 + 9x + 3x2 - 9x + 27 - x3 - 2x = 0
=> (x3 - x3) + (-3x2 + 3x2) + (9x - 9x - 2x) + 27 = 0
=> -2x + 27 = 0
=> -2x = -27
=> x = 27/2
4. \(\left(x-2\right)^3-x^3+6x^2=7\)
=> x3 - 6x2 + 12x - 8 - x3 + 6x2 = 7
=> (x3 - x3) + (-6x2 + 6x2) + 12x - 8 = 7
=> 12x - 8 = 7
=> 12x = 15
=> x = 5/4
5. \(3\left(x-2\right)^2+9\left(x-1\right)-3\left(x^2+x-3\right)=12\)
=> 3x2 - 12x + 12 + 9x - 9 - 3x2 - 3x + 9 = 12
=> (3x2 - 3x2) + (-12x + 9x - 3x) + (12 - 9 + 9) = 12
=> -6x + 12 = 12
=> -6x = 0
=> x = 0
6. \(\left(4x+3\right)^2-\left(4x-3\right)^2-5x-2=0\)
=> 48x - 5x - 2 = 0
=> 43x - 2 = 0
=> 43x = 2
=> x = 2/43
Còn bài cuối tự làm :>
Anh Sang làm cầu kì quá ;-;
1. ( x + 2 )( x2 - 2x + 4 ) - ( x3 + 2x2 ) = 5
<=> x3 + 8 - x3 - 2x2 = 5
<=> 8 - 2x2 = 5
<=> 2x2 = 3
<=> x2 = 3/2
<=> \(x^2=\left(\pm\sqrt{\frac{3}{2}}\right)^2\)
<=> \(x=\pm\sqrt{\frac{3}{2}}\)
2. ( x + 5 )2 - 6 = 0
<=> ( x + 5 )2 - ( √6 )2 = 0
<=> ( x + 5 - √6 )( x + 5 + √6 ) = 0
<=> \(\orbr{\begin{cases}x+5-\sqrt{6}=0\\x+5+\sqrt{6}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\sqrt{6}-5\\x=-\sqrt{6}-5\end{cases}}\)
3. ( x + 3 )( x2 - 3x + 9 ) - x3 = 2x
<=> x3 + 27 - x3 = 2x
<=> 27 = 2x
<=> x = 27/2
4. ( x - 2 )3 - x3 + 6x2 = 7
<=> x3 - 6x2 + 12x - 8 - x3 + 6x2 = 7
<=> 12x - 8 = 7
<=> 12x = 15
<=> x = 15/12 = 5/4
5. 3( x - 2 )2 + 9( x - 1 ) - 3( x2 + x - 3 ) = 12
<=> 3( x2 - 4x + 4 ) + 9x - 9 - 3x2 - 3x + 9 = 12
<=> 3x2 - 12x + 12 + 6x - 3x2 = 12
<=> -6x + 12 = 12
<=> -6x = 0
<=> x = 0
6. ( 4x + 3 )2 - ( 4x - 3 )2 - 5x - 2 = 0
<=> 16x2 + 24x + 9 - ( 16x2 - 24x + 9 ) - 5x - 2 = 0
<=> 16x2 + 24x + 9 - 16x2 + 24x - 9 - 5x - 2 = 0
<=> 43x - 2 = 0
<=> 43x = 2
<=> x = 2/43
7, ( 4x + 7 )( 2 - 3x ) - ( 6x + 2 )( 5 - 2x ) = 0
<=> -12x2 - 13x + 14 - ( -12x2 + 26x + 10 ) = 0
<=> -12x2 - 13x + 14 + 12x2 - 26x - 10 = 0
<=> -39x + 4 = 0
<=> -39x = -4
<=> x = 4/39
1) \(2\left(x+2\right)-\left(3x+1\right)\left(x+2\right)=0\)
\(\left(x+2\right)\left(2-3x-1\right)=0\)
\(\left(x+2\right)\left(1-3x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+2=0\\1-3x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-2\\x=\frac{1}{3}\end{cases}}}\)
2) \(3x\left(x-3\right)-\left(2x-6\right)=0\)
\(3x\left(x-3\right)-2\left(x-3\right)=0\)
\(\left(x-3\right)\left(3x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\3x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=\frac{2}{3}\end{cases}}}\)
3) \(\left(2x-1\right)^2=\left(3x-5\right)^2\)
\(\left(2x-1\right)^2-\left(3x-5\right)^2=0\)
\(\left(2x-1-3x+5\right)\left(2x-1+3x-5\right)=0\)
\(\left(4-x\right)\left(5x-6\right)=0\)
\(\Rightarrow\orbr{\begin{cases}4-x=0\\5x-6=0\end{cases}\Rightarrow\orbr{\begin{cases}x=4\\x=\frac{6}{5}\end{cases}}}\)
4) \(\left(4x+3\right)\left(x-1\right)=x^2-1\)
\(\left(4x+3\right)\left(x-1\right)=\left(x+1\right)\left(x-1\right)\)
\(\left(4x+3\right)\left(x-1\right)-\left(x+1\right)\left(x-1\right)=0\)
\(\left(x-1\right)\left(4x+3-x-1\right)=0\)
\(\left(x-1\right)\left(3x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\3x+2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=\frac{-2}{3}\end{cases}}}\)
5) \(6-4x-\left(2x-3\right)\left(x-3\right)=0\)
\(-2\left(2x-3\right)-\left(2x-3\right)\left(x-3\right)=0\)
\(\left(2x-3\right)\left(-2-x+3\right)=0\)
\(\left(2x-3\right)\left(1-x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-3=0\\1-x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=1\end{cases}}}\)
6) \(2x^2-5x-7=0\)
\(2x^2+2x-7x-7=0\)
\(2x\left(x+1\right)-7\left(x+1\right)=0\)
\(\left(x+1\right)\left(2x-7\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+1=0\\2x-7=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-1\\x=\frac{7}{2}\end{cases}}}\)
7) \(x^2-x-12=0\)
\(x^2+3x-4x-12=0\)
\(x\left(x+3\right)-4\left(x+3\right)\)
\(\left(x+3\right)\left(x-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+3=0\\x-4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-3\\x=4\end{cases}}}\)
8) \(3x^2+14x-5=0\)
\(3x^2+15x-x-5=0\)
\(3x\left(x+5\right)-\left(x+5\right)=0\)
\(\left(x+5\right)\left(3x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+5=0\\3x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-5\\x=\frac{1}{3}\end{cases}}}\)
\(d,\frac{10x+3}{8}=\frac{7-8x}{12}\)
\(\left(10x+3\right):8=\left(7-8x\right):12\)
\(\left(10x+3\right).\frac{1}{8}=\left(7-8x\right).\frac{1}{12}\)
\(\frac{5}{4}x+\frac{3}{8}=\frac{7}{12}-\frac{8}{12}x\)
\(\frac{5}{4}x+\frac{8}{12}x=\frac{7}{12}-\frac{3}{8}\)
\(\frac{23}{12}x=\frac{5}{24}\)
\(x=\frac{5}{46}\)
E mới lớp 6 nên giải sai thì thông cảm ạ UwU
\(b,\frac{x}{10}-\left(\frac{x}{30}+\frac{2x}{45}\right)=\frac{4}{5}\)
\(< =>\frac{9x}{90}-\frac{7x}{90}=\frac{4}{5}\)
\(< =>\frac{x}{45}=\frac{32}{45}\)
\(< =>x=32\)
\(d,\frac{10x+3}{8}=\frac{7-8x}{12}\)
\(< =>\left(10x+3\right).12=\left(7-8x\right).8\)
\(< =>120x+36=56-64x\)
\(< =>184x=56-36=20\)
\(< =>x=\frac{20}{184}=\frac{5}{46}\)
bài 1:
a) ĐKXĐ: x khác 0; x khác -1
\(\frac{x-1}{x}+\frac{1-2x}{x^2+x}=\frac{1}{x+1}\)
<=> \(\frac{x-1}{x}+\frac{1-2x}{x\left(x+1\right)}=\frac{1}{x+1}\)
<=> (x - 1)(x + 1) + 1 - 2x = x
<=> x^2 - 2x = x
<=> x^2 - 2x - x = 0
<=> x^2 - 3x = 0
<=> x(x - 3) = 0
<=> x = 0 hoặc x - 3 = 0
<=> x = 0 hoặc x = 0 + 3
<=> x = 0 (ktm) hoặc x = 3 (tm)
=> x = 3
b) ĐKXĐ: x khác +-3; x khác -7/2
\(\frac{13}{\left(x-3\right)\left(2x+7\right)}+\frac{1}{2x+7}=\frac{6}{x^2-9}\)
<=> \(\frac{13}{\left(x-3\right)\left(2x+7\right)}+\frac{1}{2x+7}=\frac{6}{\left(x-3\right)\left(x+3\right)}\)
<=> 13(x + 3) + (x - 3)(x + 3) = 6(2x + 7)
<=> 13x + 30 + x^2 = 12x + 42
<=> 13x + 30 + x^2 - 12x - 42 = 0
<=> x - 12 + x^2 = 0
<=> (x - 3)(x + 4) = 0
<=> x - 3 = 0 hoặc x + 4 = 0
<=> x = 0 + 3 hoặc x = 0 - 4
<=> x = 3 (ktm) hoặc x = -4 (tm)
=> x = -4
c) ĐKXĐ: x khác +-1
\(\frac{x}{x-1}-\frac{2x}{\left(x-1\right)\left(x+1\right)}=0\)
<=> x(x + 1) - 2x = 0
<=> x^2 + x - 2x = 0
<=> x^2 - x = 0
<=> x(x - 1) = 0
<=> x = 0 hoặc x - 1 = 0
<=> x = 0 hoặc x = 0 + 1
<=> x = 0 (tm) hoặc x = 1 (ktm)
=> x = 0
d) \(\frac{x^2+2x}{x^2+1}-2x=0\)
<=> \(\frac{x\left(x+2\right)}{x^2+1}-2x=0\)
<=> x(x + 2) - 2x(x^2 + 1) = 0
<=> x^2 - 2x^3 = 0
<=> x^2(1 - 2x) = 0
<=> x^2 = 0 hoặc 1 - 2x = 0
<=> x = 0 hoặc -2x = 0 - 1
<=> x = 0 hoặc -2x = -1
<=> x = 0 hoặc x = 1/2
bài 2:
(x - 1)(x^2 + 3x - 2) - (x^3 - 1) = 0
<=> x^3 + 3x^2 - 2x - x^2 - 3x + 2 - x^2 + 1 = 0
<=> 2x^2 - 2x - 3x + 3 = 0
<=> 2x(x - 1) - 3(x - 1) = 0
<=> (2x - 3)(x - 1) = 0
<=> 2x - 3 = 0 hoặc x - 1 = 0
<=> 2x = 0 + 3 hoặc x = 0 + 1
<=> 2x = 3 hoặc x = 1
<=> x = 3/2 hoặc x = 1
bài 3:
(x^3 + x^2) + (x^2 + x) = 0
<=> x^3 + x^2 + x^2 + x = 0
<=> x^3 + 2x^2 + x = 0
<=> x(x^2 + 2x + 1) = 0
<=> x(x + 1)^2 = 0
<=> x = 0 hoặc x + 1 = 0
<=> x = 0 hoặc x = 0 - 1
<=> x = 0 hoặc x = -1