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Câu hỏi của Vương Nguyễn Thanh Triều - Toán lớp 8 - Học toán với OnlineMath

Bài làm:

Vì a,b,c khác 0 nên:

Ta có: \(a\left(y+z\right)=b\left(z+x\right)=c\left(x+y\right)\)

\(\Leftrightarrow\frac{y+z}{bc}=\frac{z+x}{ca}=\frac{x+y}{ab}\)  (1) (chia cả 3 vế cho abc)

Áp dụng t/c dãy tỉ số bằng nhau ta được:
\(\left(1\right)=\frac{x+y-z-x}{ab-ca}=\frac{y+z-x-y}{bc-ab}=\frac{z+x-y-z}{ca-bc}\)

\(=\frac{y-z}{a\left(b-c\right)}=\frac{z-x}{b\left(c-a\right)}=\frac{x-y}{c\left(a-b\right)}\)

=> đpcm

Bài làm:

Vì a,b,c khác 0 nên:

Ta có: 

  (1) (chia cả 3 vế cho abc)

Áp dụng t/c dãy tỉ số bằng nhau ta được:

=> đpcm

j

a: Ta có: \(\left(ac+bd\right)^2-\left(ad+bc\right)^2\)

\(=a^2c^2+b^2d^2+2abcd-a^2d^2-b^2c^2-2abcd\)

\(=a^2\left(c^2-d^2\right)-b^2\left(c^2-d^2\right)\)

\(=\left(a^2-b^2\right)\left(c^2-d^2\right)\)

\(\text{Đặt }\dfrac{a}{x}=\dfrac{b}{y}=\dfrac{c}{z}=k \Rightarrow\left\{{}\begin{matrix}a=kx\\b=ky\\c=kz\end{matrix}\right.\\\Rightarrow\left(ax+by+cz\right)^2=\left(kx^2+ky^2+kz^2\right)^2\\ =\left(kx^2+ky^2+kz^2\right)\left(kx^2+ky^2+kz^2\right)\\ =\left(x^2+y^2+z^2\right)\left(k^2x^2+k^2y^2+k^2z^2\right) \\ =\left(x^2+y^2+z^2\right)\left(a^2+b^2+c^2\right)\left(đpcm\right)\)

\(\frac{ay-bx}{c}=\frac{cx-az}{b}=\frac{bz-cy}{a}\)

\(\Rightarrow\frac{acy-bcx}{c^2}=\frac{bcx-abz}{b^2}=\frac{abz-acy}{a^2}=\frac{0}{a^2+b^2+c^2}=0\)

\(\Rightarrow\hept{\begin{cases}ay-bx=0\\cx-az=0\\bz-cy=0\end{cases}}\)

\(\Rightarrow\left(ay-bx\right)^2+\left(cx-az\right)^2+\left(bz-ay\right)^2=0\)

\(\Rightarrow a^2y^2-2axby+b^2x^2+a^2z^2-2axcz+c^2x^2+b^2z^2-2bycz\)

\(+c^2y^2=0\)

\(\Rightarrow a^2x^2+a^2y^2+a^2z^2+b^2x^2+b^2y^2+b^2z^2+c^2x^2+c^2y^2+c^2z^2\)

\(=a^2x^2+b^2y^2+c^2z^2+2axby+2bycz+2axcz\)

\(\Rightarrow\left(x^2+y^2+z^2\right)\left(a^2+b^2+c^2\right)=\left(ax+by+cz\right)^2\)

\(\left(a^2+b^2\right)\left(x^2+y^2\right)=a^2x^2+a^2y^2+b^2x^2+b^2y^2\)(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 )

\(\left(a^2+b^2\right)+\left(x^2+y^2\right)=a^2x^2+a^2y^2+b^2x^2+b^2y^2\)

\(\left(ax+by\right)^2+\left(ay-bx\right)^2=a^2x^2+2axby+b^2y^2+a^2y^2-2aybx+b^2x^2\)

\(=a^2x^2+a^2y^2+b^2x^2+b^2y^2\)

Suy ra : \(\left(a^2+b^2\right)+\left(x^2+y^2\right)=\left(ax+by\right)^2+\left(ay+bx\right)^2\left(đpcm\right)\)

a/ \(\left(a^2-b^2\right)\left(c^2-d^2\right)=a^2c^2-a^2d^2-b^2c^2+b^2d^2\)

\(=\left(a^2c^2+2abcd+b^2d^2\right)-\left(a^2d^2+2abcd+b^2c^2\right)\)

\(=\left(ac+bd\right)^2-\left(ad+bc\right)^2\)

b/ \(x^2+y^2+z^2=xy+yz+zx\)

\(\Leftrightarrow2x^2+2y^2+2z^2=2xy+2yz+2zx\)

\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2zx+x^2\right)=0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y-z=0\\z-x=0\end{matrix}\right.\)

\(\Leftrightarrow x=y=z\)

chắc =1 đó chỉ cần đọc kĩ đề thôi

Bai 1: Ap dung BDT Bunhiacopxki ta co:

         \(ax+by+cz+2\sqrt {(ab+ac+bc)(xy+yz+xz)} \)

         \(≤ \sqrt {(a^2+b^2+c^2)(x^2+y^2+z^2)} + \sqrt {(ab+ac+bc)(xy+yz+zx)}+\sqrt {(ab+ac+bc)(xy+yz+zx)}\)

         \(≤ \sqrt {(a^2+b^2+c^2+2ab+2ac+2bc)(x^2+y^2+z^2+2xy+2yz+2zx)}\)

         \(= (a+b+c)(x+y+z)\) 

   =>  \(Q.E.D\)

Tiep bai 4:Ta co:

               BDT <=>  \((2+y^2z)(2+z^2x)(2+x^2y)≥(2+x)(2+y)(2+z)\)

    Sau khi khai trien con:   \(2(z^2x+y^2z+x^2y)+x^2z+z^2y+y^2x≥xy+yz+zx+2x+2y+2z \)

               Ap dung BDT Cosi ta co:

                                       \(z^2x+x ≥ 2zx \) <=> \(z^2x≥2zx-x\)

              Lam tuong tu ta co:  \(2(z^2x+y^2z+x^2y)≥4xy+4yz+4zx-2x-2y-2z \)(1)

                                        \(x^2z+{1\over z}≥2x \) <=> \(x^2z≥2x-xy \) (do xyz=1)

              Lam tuong tu ta co:  \(x^2z+z^2y+y^2x≥ 2y+2z+2x-xy-yz-zx\)(2)

Cong (1) voi (2) ta co:      VT\(≥ 3(xy+yz+zx)\)(*)

               Voi cach lam tuong tu ta cung duoc:  VT\(≥ 3(x+y+z) \)(**)

Tu (*) va (**) suy ra :   \(3 \)VT \(≥ 6(x+y+z)+3(xy+yz+zx) \)

                           <=>   VT \(≥ 2(x+y+z)+xy+yz+zx\)

                            =>   \(Q.E.D\)

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