\(\frac{4}{9}x\frac{3}{16}=?\)
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Đặt \(\frac{x}{3}-\frac{4}{x}=a\Rightarrow a^2=\frac{x^2}{9}+\frac{16}{x^2}-\frac{8}{3}\Rightarrow\frac{x^2}{9}+\frac{16}{x^2}=a^2+\frac{8}{3}\)
\(a^2+\frac{8}{3}=\frac{10}{3}a\Leftrightarrow3a^2-10a+8=0\Rightarrow\left[{}\begin{matrix}a=2\\a=\frac{4}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\frac{x}{3}-\frac{4}{x}=2\\\frac{x}{3}-\frac{4}{x}=\frac{4}{3}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-6x-12=0\\x^2-4x-12=0\end{matrix}\right.\)
\(\frac{3}{4}x\frac{8}{9}x\frac{15}{16}x...x\frac{9999}{10000}\)
\(=\frac{1.3}{2^2}.\frac{2.4}{3^2}.\frac{3.5}{4^2}.....\frac{99.101}{100^2}\)
\(=\frac{1.3.2.4.3.5.....99.101}{2.2.3.3.4.4.....100.100}\)
\(=\frac{1.2.3.....99}{2.3.4.....100}.\frac{3.4.5.....101}{2.3.4.....100}\)
\(=\frac{1}{100}.\frac{101}{2}=\frac{101}{200}\)
Ủng hộ mk nha,chúc bn học tốt!!!
\(\frac{9}{4}\cdot x^3+\frac{16}{3}=\frac{20}{3}\)
\(\Rightarrow\frac{9}{4}\cdot x^3=\frac{20}{3}-\frac{16}{3}\)
\(\Rightarrow\frac{9}{4}\cdot x^3=\frac{4}{3}\)
\(\Rightarrow x^3=\frac{4}{3}:\frac{9}{4}\)
\(\Rightarrow x^3=\frac{4}{3}\cdot\frac{4}{9}\)
\(\Rightarrow x^3=\frac{16}{27}\)
\(\Rightarrow x\in\varnothing\)
Không có giá trị x nào thỏa mãn
\(\frac{9}{4}x^3+\frac{16}{3}=\frac{20}{3}\)
\(\Leftrightarrow\frac{9}{4}x^3=\frac{20}{3}-\frac{16}{3}\)
\(\Leftrightarrow\frac{9}{4}x^3=\frac{4}{3}\)
\(\Leftrightarrow x^3=\frac{4}{3}\div\frac{9}{4}\)
\(\Leftrightarrow x^3=\frac{4}{3}.\frac{4}{9}\)
\(\Leftrightarrow x^3=\frac{16}{27}\)
\(\Leftrightarrow x\in\left\{\varnothing\right\}\)
Giải phương trình: \(\frac{x^2}{9}+\frac{16}{x^2}=\frac{10}{3}\left(\frac{x}{3}-\frac{4}{x}\right)\)
Điều kiện:\(x\ne0\)
Đặt \(\frac{x}{3}-\frac{4}{x}=t\).Ta có:\(t^2=\left(\frac{x}{3}-\frac{4}{x}\right)^2=\frac{x^2}{9}-2.\frac{x}{3}.\frac{4}{x}+\frac{16}{x^2}=\frac{x^2}{9}+\frac{16}{x^2}-\frac{8}{3}\)
\(\Rightarrow\frac{x^2}{9}+\frac{16}{x^2}=t^2+\frac{8}{3}\).Thay vào pt ta có:\(t^2+\frac{8}{3}=\frac{10}{3}.t\)
\(\Leftrightarrow3t^2-10t+8=0\)\(\Leftrightarrow3t^2-4t-6t+8=0\)
\(\Leftrightarrow t\left(3t-4\right)-2\left(3t-4\right)=0\)
\(\Leftrightarrow\left(t-2\right)\left(3t-4\right)=0\Rightarrow\orbr{\begin{cases}t=2\\t=\frac{4}{3}\end{cases}}\)
Với \(t=2\) thì \(\frac{x^2-12}{3x}=2\Leftrightarrow x^2-12-6x=0\)\(\Rightarrow x^2-6x+9-21=0\)
\(\Leftrightarrow\left(x-3\right)^2=21\Rightarrow\orbr{\begin{cases}x-3=\sqrt{21}\\x-3=-\sqrt{21}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\sqrt{21}+3\\x=3-\sqrt{21}\end{cases}}\)
Với \(t=\frac{4}{3}\) thì \(\frac{x^2-12}{3x}=\frac{4}{3}\Leftrightarrow x^2-4x-12=0\Leftrightarrow\left(x+2\right)\left(x-6\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=6\end{cases}}\)
Tập nghiệm của pt S=\(\left\{\sqrt{21}+3;3-\sqrt{21};-2;6\right\}\)
\(C=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot...\cdot\frac{9999}{10000}\)
\(C=\frac{1\cdot3}{2\cdot2}\cdot\frac{2\cdot4}{3\cdot3}\cdot\frac{3\cdot5}{4\cdot4}\cdot...\cdot\frac{99\cdot101}{100\cdot100}\)
\(C=\frac{1\cdot2\cdot3\cdot...\cdot99}{2\cdot3\cdot4\cdot...\cdot100}\cdot\frac{3\cdot4\cdot5\cdot...\cdot101}{2\cdot3\cdot4\cdot...\cdot100}\)
\(C=\frac{1}{100}\cdot\frac{101}{2}\)
\(C=\frac{101}{200}\)
\(C=\frac{3}{4}x\frac{8}{9}x\frac{15}{16}x......x\frac{9999}{10000}\)
\(C=\frac{1.3}{2^2}x\frac{2.4}{3^2}x\frac{3.5}{4^2}x....x\frac{99.101}{100^2}\)
\(C=\frac{1.3.2.4.3.5.......99.101}{2^2.3^2.4^2.......100^2}\)
\(C=\frac{1.2.3.......99}{2.3.4....100}x\frac{3.4.5.....101}{2.3.4......100}\)
\(C=\frac{1}{100}.\frac{101}{2}=\frac{1.101}{100.2}=\frac{101}{200}\)
Ủng hộ mk nha!!!!
\(\frac{x^2}{9}+\frac{16}{x^2}=\frac{10}{3}\left(\frac{x}{3}-\frac{4}{x}\right)\)
\(\Leftrightarrow\frac{x^2}{9}-\frac{10x}{9}+\frac{40}{3x}+\frac{16}{x^2}=0\)
\(\Leftrightarrow\frac{x^4-10x^3+120x+144}{9x^2}=0\)
\(\Leftrightarrow x^4-10x^3+120x+144=0\)
\(\Leftrightarrow x^4-6x^3-12x^2-4x^3+24x^2+48x-12x^2+72x+144=0\)
\(\Leftrightarrow x^2\left(x^2-6x-12\right)-4x\left(x^2-6x-12\right)-12\left(x^2-6x-12\right)=0\)
\(\Leftrightarrow\left(x^2-4x-12\right)\left(x^2-6x-12\right)=0\)
\(\Leftrightarrow\left(x^2+2x-6x-12\right)\left(x^2-6x-12\right)=0\)
\(\Leftrightarrow\left[x\left(x+2\right)-6\left(x+2\right)\right]\left(x^2-6x-12\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(x+2\right)\left(x^2-6x-12\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-6=0\\x+2=0\\x^2-6x-12=0\left(1\right)\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=6\\x=-2\end{array}\right.\)(tm)
\(\Delta_{\left(1\right)}=\left(-6\right)^2-\left(-4\left(1.12\right)\right)=84\)
\(\Rightarrow\)\(x_{1,2}=\frac{6\pm\sqrt{84}}{2}\) (tm)
Vậy pt có nghiệm là \(x=-2;x=6\)và \(x=\frac{6\pm\sqrt{84}}{2}\)
a) x+ \(\frac{4}{5}\)= \(\frac{4}{5}\)+ ( \(\frac{3}{7}\)+ \(\frac{3}{5}\))
x + \(\frac{4}{5}\)= \(\frac{4}{5}\)+ \(\frac{36}{35}\)
=> x = \(\frac{36}{35}\)
Vậy : x = \(\frac{36}{35}\)
b) \(\frac{4}{9}\)+\(\frac{8}{9}\)+ \(\frac{12}{9}\)+ ... + \(\frac{56}{9}\)
= \(\frac{4+8+12+...+56}{9}\)
Ta có : 4+8+12 + ... +56
Số các số hạng của dẫy số trên là : ( 56 -4) + 4 +1 = 14 ( số hạng )
=> Tổng trên = ( 4+56) x 14 : 2 = 420
=> \(\frac{4}{9}\)+ \(\frac{8}{9}\)+ ... + \(\frac{56}{9}\)= \(\frac{420}{9}\)= \(\frac{140}{3}\)
a)x=3/7+3/5=3x(1/7+1/5)=36/35
b)=(4+8+...+56)/9=(56+4)x14/9=280/3
\(\frac{4}{9}x\frac{3}{16}=\frac{4x3}{3x3x4x4}=\frac{1x1}{3x4}=\frac{1}{12}\)
\(\frac{7}{144}\)