a) \(\left(a+b\right)^2=\left(a-b\right)+4ab
\)
b) \(\left(a-b\right)^2=\left(a+b\right)^2-4ab\)
c) \(\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax-by\right)^2+\left(ay+bx\right)^2\)
Những câu hỏi liên quan
Cho \(\dfrac{bz+cy}{x\left(-ax+by+cz\right)}=\dfrac{cx+az}{y\left(ax-by+cz\right)}=\dfrac{ay+bx}{z\left(ax+by-cz\right)}\)
CMR : \(\dfrac{ay+bx}{c}=\dfrac{bz+cy}{a}=\dfrac{cx+az}{b}\)
b) \(\dfrac{x}{a\left(b^2+c^2-a^2\right)}=\dfrac{y}{b\left(a^2+c^2-b^2\right)}=\dfrac{z}{c\left(a^2+b^2-c^2\right)}\)
Bài 8.CM các hằng dẳng tức sau
1) \(\left(a+b\right)^2-\left(a-b\right)^2=4ab\)
2) \(\left(a+b\right)^2+\left(a-b\right)^2=2\left(a^2+b^2\right)\)
3) \(\left(a+b\right)^2-4ab=\left(a-b\right)^2\)
4)\(\left(a-b\right)^2+4ab=\left(a+b\right)^2\)
1. Ta có: \(\left(a+b\right)^2-\left(a-b\right)^2=\left(a+b+a-b\right)\left(a+b-a+b\right)\)
\(=2a.2b=4ab\)
=> đpcm
2. Ta có: \(\left(a+b\right)^2+\left(a-b\right)^2=a^2+2ab+b^2+a^2-2ab+b^2\)
\(=2a^2+2b^2=2\left(a^2+b^2\right)\)
=> đpcm
3. Ta có:\(\left(a+b\right)^2-4ab=a^2+2ab+b^2-4ab\)
\(=a^2-2ab+b^2=\left(a-b\right)^2\)
=> đpcm
4. Ta có: \(\left(a-b\right)^2+4ab=a^2-2ab+b^2+4ab\)
\(=a^2+2ab+b^2=\left(a+b\right)^2\)
\(a,\left(a+b\right)^2-\left(a-b\right)^2=4ab\)
\(\Leftrightarrow\left(a^2+b^2+2ab\right)-\left(a^2+b^2-2ab\right)=4ab\)
\(\Leftrightarrow a^2+b^2-a^2-b^2+2ab+2ab=4ab\)
\(\Leftrightarrow4ab=4ab\Leftrightarrow4ab-4ab=0\Leftrightarrow0=0\)(đpcm)
\(b,\left(a+b\right)^2+\left(a-b\right)^2=2\left(a^2+b^2\right)\)
\(\Leftrightarrow\left(a^2+b^2+2ab\right)+\left(a^2+b^2-2ab\right)=2\left(a^2+b^2\right)\)
\(\Leftrightarrow a^2+b^2+a^2+b^2+\left(2ab-2ab\right)=2\left(a^2+b^2\right)\)
\(\Leftrightarrow2\left(a^2+b^2\right)=2\left(a^2+b^2\right)\Leftrightarrow2\left(a^2+b^2\right)-2\left(a^2+b^2\right)=0\Leftrightarrow0=0\)(đpcm)
\(c,\left(a+b\right)^2-4ab=\left(a-b\right)^2\)
\(\Leftrightarrow\left(a^2+b^2+2ab\right)-4ab=a^2+b^2-2ab\)
\(\Leftrightarrow a^2+b^2-2ab=a^2+b^2-2ab\)
\(\Leftrightarrow\left(a-b\right)^2=\left(a-b\right)^2\Leftrightarrow\left(a-b\right)^2-\left(a-b\right)^2=0\Leftrightarrow0=0\)(đpcm)
\(d,\left(a-b\right)^2+4ab=\left(a+b\right)^2\)
\(\Leftrightarrow\left(a^2+b^2-2ab\right)+4ab=\left(a+b\right)^2\)
\(\Leftrightarrow a^2+b^2-2ab+4ab=\left(a+b\right)^2\)
\(\Leftrightarrow a^2+b^2+2ab=\left(a+b\right)^2\Leftrightarrow\left(a+b\right)^2=\left(a+b\right)^2\)
\(\Leftrightarrow\left(a+b\right)^2-\left(a+b\right)^2=0\Leftrightarrow0=0\)(đpcm)
1) \(\left(a+b\right)^2-\left(a-b\right)^2=\left(a+b-a+b\right)\left(a+b+a-b\right)\)
\(=2b.2a=4ab\)( đpcm )
2) \(\left(a+b\right)^2+\left(a-b\right)^2=a^2+2ab+b^2+a^2-2ab+b^2\)
\(=2\left(a^2+b^2\right)\)( đpcm )
3) \(\left(a+b\right)^2-4ab=a^2+2ab+b^2-4ab\)
\(=a^2-2ab+b^2=\left(a-b\right)^2\)( đpcm )
4) \(\left(a-b\right)^2+4ab=a^2-2ab+b^2+4ab\)
\(=a^2+2ab+b^2=\left(a+b\right)^2\)( đpcm )
Xem thêm câu trả lời
CMR:\(\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax-by\right)^2+\left(ay+bx\right)^2\)
Ta có:
\(\left(a^2+b^2\right)\left(x^2+y^2\right)=a^2x^2+a^2y^2+b^2x^2+b^2y^2\)
\(=a^2x^2-2abxy+b^2y^2+a^2y^2+2abxy+b^2x^2\) \(=\left(ax-by\right)^2+\left(ay+bx\right)\)
\(=vp\)
\(\Rightarrowđpcm\)
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CMR nếu a,b,c,x,y,z thỏa mãn điều kiện:
\(\frac{bz+cy}{x\left(-ax+by+cz\right)}=\frac{cx+az}{y\left(ax-by+cz\right)}=\frac{ay+bx}{z\left(ax+by-cz\right)}\)
thì \(\frac{x}{a\left(b^2+c^2-a^2\right)}=\frac{y}{b\left(a^2+c^2-b^2\right)}=\frac{z}{c\left(a^2+b^2-c^2\right)}\)
( Giả thiết các tỉ số đều có nghĩa )
CMR nếu a,b,c,x,y,z thỏa mãn :
\(\frac{bz+cy}{x\left(-ax+by+cz\right)}=\frac{cx+az}{y\left(ax-by+cz\right)}=\frac{ay+bx}{z\left(ax+by-cz\right)}\)
thì \(\frac{x}{a\left(b^2+c^2-a^2\right)}=\frac{y}{b\left(a^2+c^2-b^2\right)}=\frac{z}{c\left(a^2+b^2-c^2\right)}\)
( giả thiết các tỉ số đều có nghĩa )
CMR ( \(\left(a^2+b^2\right).\left(x^2+y^2\right)=\left(ax-by\right)^2+\left(ay-bx\right)^2\)
1. CHỨNG MINH RẰNGa) left(a+bright)^2left(a-bright)^2+4abb) left(a-bright)^2left(a+bright)^2-4abc) left(a^2+b^2right).left(x^2+y^2right)left(ax-byright)^2+left(ay+bxright)^22. CHỨNG MINH RẰNG : a b c KHI left(a+b+cright)^23left(ab+ac+bcright)3. CHO a + b + c 0 VÀ a^2+b^2+c^21Tính Ma^4+b^4+c^44. CHỨNG MINH RẰNG GIÁ TRỊ CÁC BIỂU THỨC SAU LUÔN LUÔN DƯƠNGa) x^2+x+1b) x^2-x+frac{1}{2}
Đọc tiếp
1. CHỨNG MINH RẰNG
a) \(\left(a+b\right)^2=\left(a-b\right)^2+4ab\)
b) \(\left(a-b\right)^2=\left(a+b\right)^2-4ab\)
c) \(\left(a^2+b^2\right).\left(x^2+y^2\right)=\left(ax-by\right)^2+\left(ay+bx\right)^2\)
2. CHỨNG MINH RẰNG : a = b = c KHI
\(\left(a+b+c\right)^2=3\left(ab+ac+bc\right)\)
3. CHO a + b + c = 0 VÀ \(a^2+b^2+c^2=1\)
Tính \(M=a^4+b^4+c^4\)
4. CHỨNG MINH RẰNG GIÁ TRỊ CÁC BIỂU THỨC SAU LUÔN LUÔN DƯƠNG
a) \(x^2+x+1\)
b) \(x^2-x+\frac{1}{2}\)
\(1.\)
\(a,\left(a+b\right)^2=a^2+2ab+b^2\)
\(\left(a-b\right)^2+4ab=a^2-2ab+b^2+4ab=a^2+2ab+b^2\)
\(\Rightarrow\left(a+b\right)^2=\left(a-b\right)^2+4ab\left(đpcm\right)\)
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a) \(x^2+x+1=x^2+x+\frac{1}{4}+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\)(luôn dương)
b) \(x^2-x+\frac{1}{2}=x^2-x+\frac{1}{4}+\frac{1}{4}=\left(x-\frac{1}{2}\right)^2+\frac{1}{4}>0\)(luôn dương)
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Chứng minh rằng \(\left(a^2+b^2\right).\left(x^2+y^2\right)=\left(ax+by\right)^2+\left(ay-bx\right)^2\)
\(\left(a^2+b^2\right)\left(x^2+y^2\right)=a^2x^2+a^2y^2+b^2x^2+b^2y^2\)(1)
\(\left(ax+by\right)^2+\left(ay-bx\right)^2\)
\(=a^2x^2+2axby+b^2y^2+a^2y^2-2aybx+b^2x^2\)
\(=a^2x^2+a^2y^2+b^2x^2+b^2y^2\)(2)
Từ (1) và (2) ta có \(\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax+by\right)^2+\left(ay-bx\right)^2\)( đpcm )
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\(\left(a^2+b^2\right)+\left(x^2+y^2\right)=a^2x^2+a^2y^2+b^2x^2+b^2y^2\)
\(\left(ax+by\right)^2+\left(ay-bx\right)^2=a^2x^2+2axby+b^2y^2+a^2y^2-2aybx+b^2x^2\)
\(=a^2x^2+a^2y^2+b^2x^2+b^2y^2\)
Suy ra : \(\left(a^2+b^2\right)+\left(x^2+y^2\right)=\left(ax+by\right)^2+\left(ay+bx\right)^2\left(đpcm\right)\)
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CM CÁC HẰNG ĐẲNG THỨC ;
\(\left(A^2+B^2+C^2\right)\left(X^2+Y^2+Z^2\right)=\left(AX+BY+CZ\right)^2+\left(AY-BX\right)^2+\left(AZ-CX\right)^2+\left(BZ-CY\right)^2\)
VP=\(A^2X^2+B^2Y^2+C^2Z^2+A^2Y^2+B^2X^2+A^2Z^2+C^2X^2+B^2Z^2+C^2Y^2\)
=\(A^2\left(X^2+Y^2+Z^2\right)+B^2\left(X^2+Y^2+Z^2\right)+C^2\left(X^2+Y^2+Z^2\right)\)
=\(\left(X^2+Y^2+Z^2\right)\left(A^2+B^2+C^2\right)\)
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