Tìm x, biết:
a, (2x+3)3-(x-7)2=0
b, (x2-1)3-(x4+x2+1)(x2-1)=0
c,(x-3)2-(x-3)(x2+6x+9)+6(x+1)2+3x2=-33
Giúp mk đc k?!😭☹️
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a, 3x2 - 8x + 4
= 3x2 - 6x - 2x + 4
= 3x(x - 2) - 2(x - 2)
= (3x - 2)(x - 2)
b, x2 - 4xy + 3y2
= x2 - xy - 3xy + 3y2
= x(x - y) - 3y(x - y)
= (x - 3y)(x - y)
\(a)3x^2-8x+4=3x^2-6x-2x+4=3x\left(x-2\right)-2\left(x-2\right)=\left(3x-2\right)\left(x-2\right)\)
\(b)x^2-4xy+3y^2=x^2-xy-3xy+3y^2=x\left(x-y\right)-3y\left(x-y\right)=\left(x-3y\right)\left(x-y\right)\)
\(c)2x^2+3881x-17505=2x^2+3890x-9x-17505=2x\left(x+1945\right)-9\left(x+1945\right)\)
\(=\left(2x-9\right)\left(x+1945\right)\)
\(a\left(b+c\right)^2\left(b-c\right)+b\left(c+a\right)^2\left(c-2\right)+c\left(a+b\right)^2\left(a-b\right)\)
\(=\left(b-c\right)\left(c-a\right)\left(c-b\right)\left(c+b+a\right)\)
nguồn câu hỏi tương tự
Trang 136 trong nâng cao phát triển có viết rồi mình cóp nó vô để mọi người dễ đọc nhé !
\(\left(am+bc\right)\left(bm+ac\right)\left(cm+ab\right)\)
\(=\left[a.\left(a+b+c\right)+bc\right]\left[b.\left(a+b+c\right)+ac\right]\left[c.\left(a+b+c\right)+ab\right]\)
\(=\left(a^2+ab+ac+bc\right)\left(ba+b^2+bc+ac\right)\left(ca+cb+c^2+ab\right)\)
\(=\left[\left(a^2+ab\right)+\left(ac+bc\right)\right]\left[\left(ba+b^2\right)+\left(bc+ac\right)\right]\left[\left(ca+c^2\right)\left(cb+ab\right)\right]\)
\(=\left[a\left(a+b\right)+c\left(a+b\right)\right]\left[b\left(a+b\right)+c\left(b+a\right)\right]\left[c\left(a+c\right)b\left(b+b\right)\right]\)
\(=\left(a+b\right)\left(a+c\right)\left(a+b\right)\left(b+c\right)\left(a+c\right)\left(b+c\right)\)
\(=\left(a+b\right)^2\left(a+c\right)^2\left(b+c\right)^2\)
\(\Rightarrowđpcm\)
\(\left(am+bc\right)\left(bm+ac\right)\left(cm+ab\right)\)
\(=\left[a\left(a+b+c\right)+bc\right]\left[b\left(a+b+c\right)+ac\right]\left[c\left(a+b+c\right)+ab\right]\)
\(=\left(a^2+ab+ac+bc\right)\left(ab+b^2+bc+ac\right)\left(ac+bc+c^2+ab\right)\)
\(=\left[\left(a^2+ab\right)+\left(ac+bc\right)\right]\left[\left(ab+b^2\right)+\left(bc+ac\right)\right]\left[\left(ac+c^2\right)+\left(bc+ab\right)\right]\)
\(=\left[a\left(a+b\right)+c\left(a+b\right)\right]\left[b\left(a+b\right)+c\left(a+b\right)\right]\left[c\left(a+c\right)+b\left(a+c\right)\right]\)
\(=\left(a+c\right)\left(a+b\right)\left(b+c\right)\left(a+b\right)\left(b+c\right)\left(a+c\right)\)
\(=\left(a+b\right)^2\left(a+c\right)^2\left(b+c\right)^2\)
\(\Rightarrowđpcm\)
\(4x^4-10x^3+8x^2-5x-1=0\)
\(\left(x^4-x^3+2x^2\right)-\left(4x^3-4x^2+8x\right)+\left(2x^2-2x+4\right)=0\)
\(x^2\left(x^2-x+2\right)-4x\left(x^2-x+2\right)+2\left(x^2-x+2\right)=0\)
\(\left(x^2-x+2\right)\left(x^2-4x+2\right)=0\)
\(\left[\left(x-\frac{1}{2}\right)^2+\frac{7}{4}\right]\left(x^2-4x+2\right)=0\)
Vì \(\left[\left(x-\frac{1}{2}\right)^2+\frac{7}{4}\right]>0\)\(\Rightarrow x^2-4x+2=0\)
\(\Rightarrow\left(x-2\right)^2=2\)\(\Rightarrow x-2=\pm\sqrt{2}\)
\(\Rightarrow\orbr{\begin{cases}x=\sqrt{2}+2\\x=2-\sqrt{2}\end{cases}}\)
( x - 5 )4 + ( x - 3 )4 = 16
Đặt x - 4 = a
\(\Rightarrow\)x - 5 = a -1
x - 3 = a +1
Khi đó phương trình trở thành:
( a - 1 )4 + ( a + 1 )4 = 16
\(\Leftrightarrow\)[ ( a - 1 )4 + 2.( a - 1 )2.( a + 1 )2 + ( a + 1 )4 ] - 2.( a - 1 )2.( a + 1 )2 = 16
\(\Leftrightarrow\)[ ( a - 1 )2 + ( a + 1 )2 ]2 - 2.( a - 1 )2.( a + 1 )2 = 16
\(\Leftrightarrow\)( a2 - 2a + 1 + a2 + 2a + 1 )2 - 2.( a2 - 1 )2 = 16
\(\Leftrightarrow\)( 2a2 + 2 )2 - 2.( a4 - 2a2 + 1 ) = 16
\(\Leftrightarrow\)4a4 + 8a2 + 4 - 2a4 + 4a2 - 2 - 16 = 0
\(\Leftrightarrow\) 2a4 + 12a2 - 14 = 0
\(\Leftrightarrow\)2.( a4 + 6a2 - 7 ) = 0
\(\Leftrightarrow\) a4 + 6a2 - 7 = 0
\(\Leftrightarrow\) a4 + 7a2 - a2 - 7 = 0
\(\Leftrightarrow\) a2.( a2 + 7 ) - ( a2 + 7 ) = 0
\(\Leftrightarrow\)( a2 - 1 ).( a2 + 7 ) = 0
\(\Leftrightarrow\)\(\orbr{\begin{cases}a^2-1=0\\a^2+7=0\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}a=\pm1\\a^2=-7\left(lo\text{ại}\right)\end{cases}}\)
Với a = 1 Với a = -1
\(\Rightarrow\) x - 4 = 1 \(\Rightarrow\) x - 4 = -1
\(\Leftrightarrow\) x = 5 \(\Leftrightarrow\) x = 3
Vậy x = 5 , x = 3
\(4x^2+y^2-2x-y-2xy+1=1\)
\(\Leftrightarrow4x^2-4xy+y^2-2x-y+2xy=0\)
\(\Leftrightarrow\left(2x-y\right)^2-2x-y+2xy=0\)
\(\Leftrightarrow x\left[\left(2x-y\right)-2x-y+2xy\right]=0\)
\(\Leftrightarrow x\left(2x-y\right)^2-2x^2+xy=0\)
\(\Leftrightarrow x\left[\left(2x-y\right)^2-2x+y\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\\left(2x-y\right)^2-2x+y=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\\left(2.0-y\right)^2-2.0+y=0\end{cases}}}\) (thay x=0 vào biểu thức dưới)
\(\Leftrightarrow x=0\) hoặc \(y^2+y=0\Leftrightarrow\orbr{\begin{cases}y=0\\y=-1\end{cases}}\) (mà x;y nguyên dương )=>y=0
Vậy x=0 ;y=0
\(\Leftrightarrow\orbr{\begin{cases}x=0\left(tm\right)\\y^2+y=0\Leftrightarrow\orbr{\begin{cases}y=0\left(tm\right)\\y=-1\left(ktm\right)\end{cases}}\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=0\left(tm\right)\\y^2+y=0\Leftrightarrow\orbr{\begin{cases}y=0\left(tm\right)\\y=-1\left(ktm\right)\end{cases}}\end{cases}}\)
Bạn sai rồi nhé. Khi ta giải đc x=0 ở Th1 thì không được áp dụng x=0 ở th2
\(3x\left(x+7\right)+21-3x^2=0\)
\(\Leftrightarrow3x^2+21x+21-3x^2=0\)
\(\Leftrightarrow21x+21=0\)
\(\Leftrightarrow21\left(x+1\right)=0\)
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=-1\)
\(5x\left(3x+7\right)-15x^2=70\)
\(\Leftrightarrow15x^2+35x-15x^2=70\)
\(\Leftrightarrow35x=70\)
\(\Leftrightarrow x=2\)