LÀM GÌ ĐỂ XÓA CÂU HỎI NÀY

ta có: ab=2; ac+ bd = 2

=> ab+cd=2=>2-ab=cd=1

vậy 1-cd=0 thì ko phải là số âm

cd=ac.bd≤\(\dfrac{\left(ac+bd\right)^2}{4}=1\)

Ta gọi tổng trên là B.

Ta luôn có : \(\frac{1}{2^2}>0\)

\(\Rightarrow B>0\left(1\right)\)

Ta có : 

\(B< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{44.45}\)

\(B< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{44}-\frac{1}{45}\)

\(B< 1-\frac{1}{45}\)

\(B< \frac{44}{45}\)

\(\Rightarrow B< 1\left(2\right)\)

Từ (1) và (2 ) 

\(\Rightarrow0< B< 1\)

=> Tổng B không nguyên

\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{45^2}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{45^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{44.45}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{45^2}< \frac{44}{45}< 1\)

\(\Leftrightarrow0< \frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{45^2}< 1\)

=> Tổng trên không nguyên

\(A\left(x\right)+B\left(x\right)=3x^4-\frac{3}{4}x^3+2x^2-3+8x^4+\frac{1}{5}x^3-9x+\frac{2}{5}\)

\(=11x^4-\frac{11}{20}x^3+2x^2-\frac{13}{5}-9x\)

\(A\left(x\right)-B\left(x\right)=3x^4-\frac{3}{4}x^3+2x^2-3-8x^4-\frac{1}{5}x^3+9x-\frac{2}{5}\)

\(=-5x^4-\frac{19}{20}x^3+2x^2-\frac{17}{5}+9x\)

Bn làm nót nhé, tương tự thôi 

\(A\left(x\right)+B\left(x\right)\)

\(=\left(3x^4-\frac{3}{4}x^3+2x^2-3\right)+\left(8x^4+\frac{1}{5}x^3-9x+\frac{2}{5}\right)\)

\(=11x^4-\frac{11}{20}x^3+2x^2-9x-\frac{13}{5}\)

\(A\left(x\right)-B\left(x\right)\)

\(=3x^4-\frac{3}{4}x^3+2x^2-3-8x^4-\frac{1}{5}x^3+9x-\frac{2}{5}\)

\(=-5x^4-\frac{19}{20}x^3+2x^2+9x-\frac{17}{5}\)

\(B\left(x\right)-A\left(x\right)\)

\(=8x^4+\frac{1}{5}x^3-9x+\frac{2}{5}-3x^4+\frac{3}{4}x^3+2x^2-3\)

\(=5x^4+\frac{19}{20}x^3+2x^2-9x-\frac{13}{5}\)

Từ \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a\cdot b}{c\cdot d}=\frac{a^2-b^2}{c^2-d^2}\left(đpcm\right)\)

đặt \(\frac{a}{b}\)=\(\frac{c}{d}\)=k =>a=bk; c=dk

xét: \(\frac{ab}{cd}\)=\(\frac{bk.b}{dk.d}\)=\(\frac{b^2}{d^2}\)

\(\frac{a^2-b^2}{c^2-d^2}\)=\(\frac{b^2k^2-b^2}{d^2k^2-d^2}\)=\(\frac{b^2\left(k^2-1\right)}{d^2\left(k^2-1\right)}\)=\(\frac{b^2}{d^2}\)

=> \(\frac{ab}{cd}\)=\(\frac{a^2-b^2}{c^2-d^2}\)đpcm

tương tự

xét:  \(\left(\frac{a+b}{c+d}\right)^2\)=\(\left(\frac{bk+b}{dk+d}\right)^2\)=\(\left(\frac{b\left(k+1\right)}{d\left(k+1\right)}\right)^2\)=\(\frac{b^2}{d^2}\)

\(\frac{a^2+b^2}{c^2+d^2}\)=\(\frac{b^2k^2+b^2}{d^2k^2+d^2}\)=\(\frac{b^2\left(k+1\right)}{d^2\left(k+1\right)}\)=\(\frac{b^2}{d^2}\)

=> \(\left(\frac{a+b}{c+d}\right)^2\)=\(\frac{a^2+b^2}{c^2+d^2}\)đpcm

a) \(2x-3=0\Leftrightarrow x=\frac{3}{2}\)

b) \(x^3+1=0\)

\(\Leftrightarrow x^3=-1\)

\(\Leftrightarrow x^3=\left(-1\right)^3\Leftrightarrow x=-1\)

c) \(x^2-2x+1=0\)

\(\Leftrightarrow\left(x-1\right)^2=0\)

\(\Leftrightarrow x-1=0\)

\(\Leftrightarrow x=1\)

a) \(2x-3=0\)

\(\Leftrightarrow2x=0+3\)

\(\Leftrightarrow2x=3\)

\(\Leftrightarrow x=3\div2\)

\(\Leftrightarrow x=1,5\)

Vậy nghiệm của đa thức 2x - 3 là 1,5.

b. \(x^3+1=0\)

\(\Leftrightarrow x^3=0-1\)

\(\Leftrightarrow x^3=-1\)

\(\Leftrightarrow x^3=-1^3\)

\(\Leftrightarrow x=-1\)

Vậy nghiệm của đa thức x³ + 1 là - 1.

c) \(x^2-2x+1=0\)

\(\Leftrightarrow x^2-2x=0-1\)

\(\Leftrightarrow x\left(x-2\right)=-1=-1.1=1.\left(-1\right)\)

Vậy đa thức \(x^2-2x+1\) vô nghiệm