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Có \(\hept{\begin{cases}\left|a\right|+\left|b\right|\ge0\\\left|a-b\right|\ge0\end{cases}}\)
\(\left|a\right|+\left|b\right|\ge\left|a-b\right|\)
\(\Leftrightarrow\left(\left|a\right|+\left|b\right|\right)^2\ge\left|a-b\right|^2\)
\(\Leftrightarrow a^2+2.\left|a\right|.\left|b\right|+b^2\ge a^2-2ab+b^2\)
\(\Leftrightarrow2.\left|a\right|.\left|b\right|\ge2ab\)( luôn đúng )
\(\Rightarrow\left|a\right|+\left|b\right|\ge\left|a-b\right|\)
đpcm
Gải sử..
\(1)\)\(\left|a\right|+\left|b\right|\ge\left|a-b\right|\)
\(\Leftrightarrow\)\(\left(\left|a\right|+\left|b\right|\right)^2\ge\left|a-b\right|^2\)
Có \(\left|a-b\right|^2=\left(a-b\right)^2\)
\(\Leftrightarrow\)\(a^2+2\left|ab\right|+b^2\ge a^2-2ab+b^2\)
\(\Leftrightarrow\)\(\left|ab\right|\ge-ab\) ( đúng )
Dấu "=" xảy ra \(\Leftrightarrow\)\(ab< 0\)
\(2)\)\(\left|a\right|+\left|b\right|+\left|c\right|\ge\left|a+b+c\right|\)
\(\Leftrightarrow\)\(\left(\left|a\right|+\left|b\right|+\left|c\right|\right)^2\ge\left|a+b+c\right|^2\)
Có \(\left|a+b+c\right|^2=\left(a+b+c\right)^2\)
\(\Leftrightarrow\)\(a^2+b^2+c^2+2\left|ab\right|+2\left|bc\right|+2\left|ca\right|\ge a^2+b^2+c^2+2ab+2bc+2ca\)
\(\Leftrightarrow\)\(\left|ab\right|+\left|bc\right|+\left|ca\right|\ge ab+bc+ca\) ( đúng )
Dấu "=" xảy ra khi a, b, c cùng dấu ( cùng dương hoặc cùng âm )
\(3)\) Sai đề thì phải. Giả sử \(a=3;b=0\) thì \(\left|a+b\right|< \left|1+ab\right|\)
\(\Leftrightarrow\)\(\left|3+0\right|< \left|1+3.0\right|\)\(\Leftrightarrow\)\(3< 1\) ( ??? )
...
1. Tìm x:
a) \(\left(x+36\right)^2=1936\Leftrightarrow x+36=\pm44.\) Vậy x = 8 hoặc x = -80
b) \(\left(\dfrac{3}{5}\right)^{x+2}=\dfrac{81}{625}\Leftrightarrow\left(\dfrac{3}{5}\right)^{x+2}=\left(\dfrac{3}{5}\right)^4\Leftrightarrow x+2=4\Leftrightarrow x=2\)
c) Xem lại đề
d) \(\left(\dfrac{9}{16}\right)^{x-5}=\left(\dfrac{4}{3}\right)^4\Leftrightarrow\left(\dfrac{3}{4}\right)^{2\left(x-5\right)}=\left(\dfrac{3}{4}\right)^{-4}\Leftrightarrow2\left(x-5\right)=-4\Leftrightarrow x=3\)
e) \(\left(\dfrac{3}{5}\right)^x.\left(\dfrac{125}{27}\right)^x=\dfrac{81}{625}\Leftrightarrow\left(\dfrac{3}{5}.\dfrac{125}{27}\right)^x=\left(\dfrac{3}{5}\right)^4\Leftrightarrow\left(\dfrac{5}{3}\right)^{2x}=\left(\dfrac{5}{3}\right)^{-4}\Leftrightarrow2x=-4\) Vậy x = -2
3. Tính giá trị của biểu thức:
\(A=\left\{-\left[\left(\dfrac{1}{x}\right)^2\right]^3\right\}^5.\left\{-\left[\left(-x\right)^5\right]^2\right\}^3\) \(\left(x\notin0\right)\)
\(=\left\{-\left[-\dfrac{1}{x^2}\right]^3\right\}^5.\left\{-\left[-\left(-x\right)^5\right]^2\right\}^3=\left\{-\left[-\dfrac{1}{x^6}\right]\right\}^5.\left\{-\left[x^5\right]^2\right\}^3\)
\(=\left\{\dfrac{1}{x^6}\right\}^5.\left\{-x^{10}\right\}^3=\dfrac{1}{x^{30}}.\left(-x^{30}\right)=-1\)
Ta có:\(\left|a\right|,\left|b\right|\) \(\leq\) \(1\)
\(\implies\) \(\left(1-a\right).\left(1-b\right)\) \(\geq\) \(0\)
\(\implies\) \(1-b-a+ab\)\(\geq\) \(0\)
\(\implies\) \(1+ab\) \(\geq\) \(a+b\)
\(\implies\) \(\left|1+ab\right|\) \(\geq\) \(\left|a+b\right|\) \(\left(đpcm\right)\)
\(x^2-y^2\)
\(=x^2-xy+xy-y^2=x.\left(x-y\right)+y.\left(x-y\right)=\left(x+y\right).\left(x-y\right)\)
\(\left(x+y\right).\left(x^2-xy+y^2\right)\)
\(=x^3-x^2y+xy^2+x^2y-xy^2+y^3=x^3+y^3\)
Thực hiện phép tính ở VP ta có:
a) \(\dfrac{1}{a}-\dfrac{1}{a+1}=\dfrac{a+1}{a\left(a+1\right)}-\dfrac{a}{a\left(a+1\right)}=\dfrac{1}{a\left(a+1\right)}\)
VP bằng VT nên đẳng thức trên là đúng
b) \(\dfrac{1}{a\left(a+1\right)}-\dfrac{1}{\left(a+1\right)\left(a+2\right)}=\dfrac{a+2}{a\left(a+1\right)\left(a+2\right)}-\dfrac{a}{a\left(a+1\right)\left(a+2\right)}\)
\(=\dfrac{2}{a\left(a+1\right)\left(a+2\right)}\)
VP bằng VT nên đẳng thức trên là đúng
a, \(\dfrac{1}{a.\left(a+1\right)}=\dfrac{1}{a}-\dfrac{1}{a+1}\)
Ta có:
\(VP=\dfrac{1}{a}-\dfrac{1}{a+1}=\dfrac{\left(a+1\right)-a}{a\left(a+1\right)}=\dfrac{1}{a\left(a+1\right)}=VT\)
\(\rightarrow\) đpcm
b, \(\dfrac{2}{a\left(a+1\right)\left(a+2\right)}=\dfrac{1}{a\left(a+1\right)}-\dfrac{1}{\left(a+1\right)\left(a+2\right)}\)
Ta có:
\(VP=\dfrac{1}{a\left(a+1\right)}-\dfrac{1}{\left(a+1\right)\left(a+2\right)}=\dfrac{1}{a}-\dfrac{1}{a+1}-\dfrac{1}{a+1}+\dfrac{1}{a+2}\)(áp dụng câu a)
\(=\dfrac{1}{a}-\dfrac{2}{a+1}+\dfrac{1}{a+2}\)
\(=\dfrac{\left(a+1\right)-2a}{a\left(a+1\right)}+\dfrac{1}{a+2}=\dfrac{\left[\left(a+1\right)-2a\right]\left(a+2\right)}{a.\left(a+1\right)\left(a+2\right)}\)
\(=\dfrac{\left(1-a\right)\left(a+2\right)+a\left(a+1\right)}{a\left(a+1\right)\left(a+2\right)}=\dfrac{a+2-a^2-2a+a^2+a}{a\left(a+1\right)\left(a+2\right)}\)
\(=\dfrac{2}{a\left(a+1\right)\left(a+2\right)}=VT\)
Chúc bạn học tốt!!!
a: Đặt f(x)=0
\(\Leftrightarrow x-2x^2+2x^2-x+4=0\)
=>4=0(loại)
b: Đặt g(x)=0
\(\Leftrightarrow x^2-5x-x^2-2x+7x=0\)
=>0x=0(luôn đúng)
c: Đặt H(x)=0
\(\Leftrightarrow x^2-x+1=0\)
Δ=1-4=-3<0
DO đó: PTVN