2.Cho đa thức R(x)=x2-2x.Tính giá trị của biểu thức
\(S=\dfrac{1}{R\left(3\right)}+\dfrac{1}{R\left(4\right)}+\dfrac{1}{R\left(5\right)}+...+\dfrac{1}{R\left(2022\right)}+\dfrac{1}{2.2023}\)
Giaỉ chi tiết
Cho đa thức R(x)=\(x^2+2x\). Tính giá trị của biểu thức
\(S=\dfrac{1}{R\left(3\right)}+\dfrac{1}{R\left(4\right)}+\dfrac{1}{R\left(5\right)}+...+\dfrac{1}{R\left(2023\right)}+\dfrac{1}{2.2023}\)
\(\dfrac{1}{R\left(x\right)}=\dfrac{1}{x\left(x+2\right)}=\dfrac{1}{2}\left(\dfrac{1}{x}-\dfrac{1}{x+2}\right)\)
\(\Rightarrow S=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2022}-\dfrac{1}{2024}+\dfrac{1}{2023}-\dfrac{1}{2025}\right)+\dfrac{1}{2.2023}\)
\(=\dfrac{1}{2}\left(\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{2024}-\dfrac{1}{2025}\right)+\dfrac{1}{2.2023}\)
Một kết quả rất xấu
Bài 1: Rút gọn biểu thức : \(\left(2-\sqrt{2}\right).\left(-5\sqrt{2}\right)-\left(3\sqrt{2}-5\right)^2\)
Bài 2: Cho biểu thức: P = \(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{2+5\sqrt{x}}{4-x}\)
a) Rút gọn biểu thức (tìm đk)
b) Tìm x để P = 2
BÀi 3: Q = \(\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
a) Rút gọn biểu thức (tìm đk)
b) Tìm giá trị của a để Q > 0
Bài 1 : Rút gọn biểu thức :
\(\left(2-\sqrt{2}\right)\left(-5\sqrt{2}\right)-\left(3\sqrt{2}-5\right)^2\)
\(=\left(-10\sqrt{2}+10\right)-\left(18-30\sqrt{2}+25\right)\)
\(=\left(-10\sqrt{2}+10\right)-\left(7-30\sqrt{2}\right)\)
\(=-10\sqrt{2}+10-7+30\sqrt{2}\)
\(=20\sqrt{2}+3\)
Bài 2:
a) ĐKXĐ : x # 4 ; x # - 4
P = \(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{2+5\sqrt{x}}{4-x}\)
P =\(\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{2+5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
P = \(\dfrac{x+2\sqrt{x}+\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
P = \(\dfrac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
P = \(\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)
b ) Để P = 2 \(\Leftrightarrow\dfrac{3\sqrt{x}}{\sqrt{x}+2}\) = 2
\(\Leftrightarrow3\sqrt{x}=2\sqrt{x}+4\)
\(\Leftrightarrow\sqrt{x}=4\)
\(\Leftrightarrow x=16\)
Vậy, để P = 2 thì x = 16.
Bài 3 :
a) ĐKXĐ : a # 1 ; a # 0, a # 4
\(Q=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(Q=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(Q=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{1}\)
\(Q=\dfrac{\sqrt{a}-2}{\sqrt{a}}\)
b) Để \(Q>0\) thì \(\dfrac{\sqrt{a}-2}{\sqrt{a}}>0\)
\(\Leftrightarrow\sqrt{a}-2>0\)
\(\Leftrightarrow\sqrt{a}>2\Leftrightarrow a>4\)
Vậy, để Q > 0 thì a > 4
\(\dfrac{\left(\dfrac{-1}{2}\right)^3-\left(\dfrac{3}{4}\right)^3.\left(-2\right)^2}{2.\left(-1\right)^5+\left(\dfrac{3}{4}\right)^2-\dfrac{3}{8}}\)
Rút gọn biểu thức trên
\(\dfrac{\left(\dfrac{-1}{2}\right)^3-\left(\dfrac{3}{4}\right)^3.\left(-2\right)^2}{2.\left(-1\right)^5+\left(\dfrac{3}{4}\right)^2-\dfrac{3}{8}}\)
\(=\dfrac{\dfrac{-1}{8}-\dfrac{27}{64}.4}{-2+\dfrac{9}{16}-\dfrac{3}{8}}=\dfrac{-\dfrac{1}{8}-\dfrac{27}{16}}{-\dfrac{29}{16}}\)
\(=\dfrac{-\dfrac{29}{16}}{-\dfrac{29}{16}}=1\)
Chúc bạn học tốt!!!
Rút gọn biểu thức D=\(\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{4}\right)....\left(1-\dfrac{1}{2015}\right)\)
\(D=\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{4}\right)....\left(1-\dfrac{1}{2015}\right)\)
\(D=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{2014}{2015}=\dfrac{1.2.3....2014}{2.3.4....2015}\)
\(D=\dfrac{1}{2015}\)
\(D=\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right)...\left(1-\dfrac{1}{2015}\right)\)
\(D=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{2014}{2015}=\dfrac{1.2.3...2014}{2.3.4...2015}\)
\(D=\dfrac{1}{2015}\)
BT18: Cho\(P\left(x\right)=5x^2+5x-4\) , \(Q\left(x\right)=2x^2-3x+1\) và \(R\left(x\right)=4x^2-x+3\)
Tính P(x)+Q(x)-R(x) rồi tính giá trị của đa thức tại \(x=-\dfrac{1}{2}\)
`@` `\text {Ans}`
`\downarrow`
`P(x)+Q(x)-R(x)`
`= 5x^2 + 5x - 4 +2x^2 - 3x + 1 - (4x^2 - x + 3)`
`= 5x^2 + 5x - 4 + 2x^2 - 3x + 1 - 4x^2 + x - 3`
`= (5x^2 + 2x^2 - 4x^2) + (5x - 3x + x) + (-4 + 1 - 3)`
`= 3x^2 + 3x - 6`
Thay `x=-1/2`
`3*(-1/2)^2 + 3*(-1/2) - 6`
`= 3*1/4 - 3/2 - 6`
`= 3/4 - 3/2 - 6`
`= -3/4 - 6 = -27/4`
Vậy, khi `x=-1/2` thì GTr của đa thức là `-27/4`
P(x)+Q(x)-R(x)
=5x^2+5x-4+2x^2-3x+1-4x^2+x-3
=2x^2+3x-6(1)
Khi x=-1/2 thì (1) sẽ là 2*1/4+3*(-1/2)-6=1/2-3/2-6=-7
1. Rút gon biểu thức chứa căn
\(\left(1+\sqrt{2}+\sqrt{3}\right)\left(1+\sqrt{2}-\sqrt{3}\right)\)
2. Cho biểu thức \(P=\left(1+\dfrac{1}{\sqrt{x-1}}\right).\dfrac{1}{x-\sqrt{x}}\)
a) Tìm ĐKXĐ và rút gọn P
b) Tìm x để \(P.\sqrt{5+2\sqrt{6}}.\left(\sqrt{x}-1\right)^2=x-2005+\sqrt{2}+\sqrt{3}\)
3. Cho biểu thức \(A=\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+1}\right)\left(1+\dfrac{1}{\sqrt{x}}\right)\)
a) Tìm ĐKXĐ và rút gọn A
b) Tìm giá trị của x để \(\sqrt{A}>A\)
1. \(\left(1+\sqrt{2}+\sqrt{3}\right)\left(1+\sqrt{2}-\sqrt{3}\right)\)
\(=\left(1+\sqrt{2}\right)^2-\sqrt{3}^2\)
\(=1+2\sqrt{2}+2-3\)
\(=2\sqrt{2}\)
3. \(A=\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+1}\right)\cdot\left(1+\dfrac{1}{\sqrt{x}}\right)\)(1)
ĐKXĐ \(x>0,x\ne1\)
pt (1) <=> \(\left(\dfrac{\sqrt{x}+1+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)}\right)\cdot\left(\dfrac{\sqrt{x}+1}{\sqrt{x}}\right)\)
\(\Leftrightarrow\dfrac{\left(\sqrt{x}+1\right)\cdot\left(\sqrt{x}+1+\sqrt{x}-1\right)}{\sqrt{x}\cdot\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)}\)
\(\Leftrightarrow\dfrac{2\sqrt{x}}{x-\sqrt{x}}\)
\(\Leftrightarrow\dfrac{\sqrt{x}\cdot2}{\sqrt{x}\cdot\left(\sqrt{x}-1\right)}\)
\(\Leftrightarrow\dfrac{2}{\sqrt{x}-1}\)
b) Để \(\sqrt{A}>A\Leftrightarrow\sqrt{\dfrac{2}{\sqrt{x}-1}}>\dfrac{2}{\sqrt{x}-1}\)
\(\Leftrightarrow\dfrac{2}{\sqrt{x}-1}>\dfrac{4}{x-2\sqrt{x}+1}\)
\(\Leftrightarrow\dfrac{2}{\sqrt{x}-1}-\dfrac{4}{x-2\sqrt{x}+1}>0\)
\(\Leftrightarrow\dfrac{2\cdot\left(\sqrt{x}-1\right)-4}{x-2\sqrt{x}+1}>0\)
\(\Leftrightarrow\dfrac{2\sqrt{2}-2-4}{x-2\sqrt{x}+1}>0\)
\(\Leftrightarrow\dfrac{2\sqrt{2}-6}{x-2\sqrt{x}+1}>0\)
Vì \(2\sqrt{2}-6< 0\Rightarrow x-2\sqrt{x}+1< 0\)
mà \(x-2\sqrt{x}+1=\left(\sqrt{x}-1\right)^2\ge0\forall x\)
Vậy không có giá trị nào của x thỏa mãn \(\sqrt{A}>A\)
(P/s Đề câu b bị sai hay sao vậy, chả có số nào mà \(\sqrt{A}>A\) cả, check lại đề giùm với nhé)
1)Đặt:
\(THANGDZ=\left(1+\sqrt{2}+\sqrt{3}\right)\left(1+\sqrt{2}-\sqrt{3}\right)\)
\(THANGDZ=\left(1+\sqrt{2}\right)^2-3\)
\(THANGDZ=1+2\sqrt{2}+2-3\)
\(THANGDZ=2\sqrt{2}\)
Thông cảm-Trình em có hạn
Rút gọn biểu thức sau:
B=\(\dfrac{\left(\dfrac{2}{3}\right)^3\times\left(-\dfrac{3}{4}\right)^2\times\left(-1\right)^5}{\left(\dfrac{2}{5}\right)^2\times\left(-\dfrac{5}{12}\right)^3}\)
Ta có:
\(B=\dfrac{\left(\dfrac{2}{3}\right)^3.\left(-\dfrac{3}{4}\right)^2.\left(-1\right)^5}{\left(\dfrac{2}{5}\right)^2.\left(-\dfrac{5}{12}\right)^3}=\dfrac{\dfrac{8}{27}.\dfrac{9}{16}.\left(-1\right)}{\dfrac{4}{25}.\left(-\dfrac{125}{1728}\right)}\\ =\dfrac{-\dfrac{1}{6}}{-\dfrac{5}{432}}=\dfrac{72}{5}\)
Vậy B = \(\dfrac{72}{5}\)
Rút gọn A=\(\left(\dfrac{3}{2}-\dfrac{2}{5}+\dfrac{1}{10}\right)+\left(\dfrac{3}{2}-\dfrac{2}{3}+\dfrac{1}{12}\right)\)
Tìm giá trị nhở nhất của biểu thức P=\(\left|x-2012\right|+\left|x-2013\right|\) với x là STN
1: \(A=\dfrac{15-4+1}{10}+\dfrac{18-8+1}{12}\)
\(=\dfrac{12}{10}+\dfrac{11}{12}\)
\(=\dfrac{6}{5}+\dfrac{11}{12}=\dfrac{72+55}{60}=\dfrac{127}{60}\)
Rút gọn biểu thức sau:
\(A=\left(1+\dfrac{2}{1.4}\right).\left(1+\dfrac{2}{2.5}\right).\left(1+\dfrac{2}{3.6}\right).....\left(1+\dfrac{2}{x\left(x+3\right)}\right)\)
\(A=\left(\dfrac{6}{1.4}\right)\left(\dfrac{12}{2.5}\right)\left(\dfrac{20}{3.6}\right)\left(\dfrac{x^2+3x+2}{x\left(x+3\right)}\right)\)
\(A=\dfrac{2.3}{1.4}.\dfrac{3.4}{2.5}.\dfrac{4.5}{3.6}...\dfrac{\left(x+1\right)\left(x+2\right)}{x\left(x+3\right)}\)
\(A=\dfrac{2.3.4...\left(x+1\right)}{1.2.3...x}.\dfrac{3.4.5...\left(x+2\right)}{4.5.6...\left(x+3\right)}=\left(x+1\right)\dfrac{3}{x+3}=\dfrac{3\left(x+1\right)}{x+3}\)