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\(M=\frac{-2}{7}x^4y\cdot\left(-\frac{21}{10}\right)xy^2z^2=\left(-\frac{2}{7}\cdot-\frac{21}{10}\right)\left(x^4x\right)\left(yy^2\right)z^2=\frac{3}{5}x^5y^3z^2\)
Hệ số 3/5
\(N=-16x^2y^2z^4\cdot\left(-\frac{1}{4}\right)xy^2z=\left(-16\cdot-\frac{1}{4}\right)\left(x^2x\right)\left(y^2y^2\right)\left(z^4z\right)=4x^3y^4z^5\)
Hệ số 4
Làm nốt b Quỳnh đag lm dở.
Ta có \(P\left(x\right)=C\left(x\right)+D\left(x\right)\)
\(P\left(x\right)=2x^4+2x-6x^2-x^3-3+4x^2+x^3-2x^2-2x^4-2x+5x^2+1\)
\(P\left(x\right)=x^2-2\)
Ta có : \(P\left(x\right)=x^2-2=0\)
\(\Leftrightarrow x^2=2\Leftrightarrow x=\pm\sqrt{2}\)
a) Thay x = \(\sqrt{2}\)vào biểu thức ta có :
\(A=\left(\sqrt{2}+1\right)\left[\left(\sqrt{2}\right)^2-2\right]=\left(\sqrt{2}+1\right).\left(2-2\right)=0\)
Giá trị của A khi x = \(\sqrt{2}\)là 0
b) Ta có \(B=\frac{2x^23x-2}{x+2}=\frac{6x^3-2}{x+2}\)
Thay x = 3 vào B ta có : \(B=\frac{6.3^3-2}{3+2}=\frac{160}{5}=32\)
Giá trị của B khi x = 3 là 32
d) Đặt \(\frac{x}{3}=\frac{y}{5}=k\Rightarrow x=3k;y=5k\)
Khi đó D = \(\frac{5\left(3k\right)^2+3.\left(5k\right)^2}{10\left(3k\right)^2-3\left(5k\right)^2}=\frac{45k^2+75k^2}{90k^2-75k^2}=\frac{120k^2}{15k^2}=8\)
=> D = 8
e) E = \(\left(1+\frac{z}{x}\right)\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)=\frac{x+z}{x}.\frac{x+y}{y}.\frac{y+z}{z}=\frac{\left(x+y\right)\left(x+z\right)\left(y+z\right)}{xyz}\)
Lại có x + y + z = 0
=> x + y = -z
=> x + z = - y
=> y + z = - x
Khi đó E = \(\frac{-xyz}{xyz}=-1\)
\(\left(a^5b^2xy^2z^{n-1}\right)\left(-\frac{5}{3}ax^5y^2z\right)^3=-\frac{125}{27}.a^8b^2x^{16}y^7z^{n+2}\)
Hệ số \(\frac{-125}{27}\)
Biến : a8b2x16y7zn + 2
\(\frac{3x-2y}{37}=\frac{5y-3z}{15}=\frac{2z-5x}{2}=\)
\(\frac{3xz-2yz}{37z}=\frac{5yx-3zx}{15x}=\frac{2zy-5xy}{2y}=\frac{3xz-2yz+5yx-3zx+2zy-5xy}{37z+15x+2y}=0\)(t/c dãy tỉ số bằng nhau)
\(\frac{3x-2y}{37}=0\Rightarrow3x=2y\Rightarrow\frac{x}{2}=\frac{y}{3}\left(1\right)\)
\(\frac{5y-3z}{15}=0\Rightarrow5y=3z\Rightarrow\frac{z}{5}=\frac{y}{3}\left(2\right)\)
\(\frac{2z-5x}{2}=0\Rightarrow2z=5x\Rightarrow\frac{x}{2}=\frac{z}{5}\left(3\right)\)
\(\Rightarrow\frac{x}{2}=\frac{y}{3}=\frac{z}{5}=\frac{10x}{20}=\frac{3y}{9}=\frac{2z}{10}=\frac{10x-3y-2z}{20-9-10}=\frac{-4}{1}=-4\)
\(x=-8,y=-12,z=-20\)
Câu 1:
\(3\left(x-1\right)=2\left(y-2\right)\Leftrightarrow3x-3=2y-4\Leftrightarrow3x=2y-1\)
\(4\left(y-2\right)=3\left(z-3\right)\Leftrightarrow4y-8=3z-9\Leftrightarrow4y=3z-1\)
Lại có:
\(3x=2y-1\Leftrightarrow6x=4y-2=3z-1-2=3z-3\)
\(\Rightarrow6x=4y-2=3z-3\)
\(\Rightarrow6x=3z-3\Leftrightarrow2x=z-1\)
\(\Rightarrow2x+3y-z=z-1+3y-z=3y-1=50\Leftrightarrow3y=51\Leftrightarrow y=17\)\(\Rightarrow\left\{{}\begin{matrix}x=11\\z=23\end{matrix}\right.\)
Câu 3:
\(\frac{a}{b}=\frac{8}{5}\Leftrightarrow\frac{a}{8}=\frac{b}{5}\Leftrightarrow\frac{1}{2}.\frac{a}{8}=\frac{1}{2}.\frac{b}{5}\Leftrightarrow\frac{a}{16}=\frac{b}{10}\) (1)
\(\frac{b}{c}=\frac{2}{7}\Leftrightarrow\frac{b}{2}=\frac{c}{7}\Leftrightarrow\frac{1}{5}.\frac{b}{2}=\frac{1}{5}.\frac{c}{7}\Leftrightarrow\frac{b}{10}=\frac{c}{35}\) (2)
Từ (1) và (2)
\(\Rightarrow\frac{a}{16}=\frac{b}{10}=\frac{c}{35}=k\)\(\Rightarrow\left\{{}\begin{matrix}a=16k\\b=10k\\c=35k\end{matrix}\right.\)
\(\Rightarrow a+b+c=16k+10k+35k=61k=61\Rightarrow k=1\)
\(\Rightarrow\left\{{}\begin{matrix}a=16k=16\\b=10k=10\\c=35k=35\end{matrix}\right.\)
Đỗ Đức Đạt
\(x=\frac{4}{5};x=-\frac{5}{6};x=\frac{7}{15}\)
\(M+N=9\left(xz\right)^2+14\left(yz\right)^2=z^2\left(9x^2+14y^2\right)\)
Câu 1 : \(x-\frac{1}{5}=\frac{3}{5}\)
\(x=\frac{3}{5}+\frac{1}{5}\)
\(x=\frac{4}{5}\)
\(\frac{1}{2}+x=-\frac{1}{3}\)
\(x=-\frac{1}{3}-\frac{1}{2}\)
\(x=-\frac{5}{6}\)
\(-x+\frac{1}{3}=-\frac{2}{15}\)
\(x=\frac{1}{3}+\frac{2}{15}\)
\(x=\frac{7}{15}\)
Câu 2 : Ta có : \(M+N=5x^2z^2+5y^2z+8x+4x^2z^2+9y^2z-8x\)
\(=9x^2x^2+14y^2z\)