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1. 2n-3 ⋮ n+1
⇒2n+2-5 ⋮ n+1
⇒2(n+1)-5 ⋮ n+1
Do n∈Z
⇒n+1 ∈ Ư(-5)={-1,1,-5,5}
⇒\(\left[{}\begin{matrix}n-1=-1\\n-1=1\\n-1=-5\\n-1=5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}n=0\\n=2\\n=-4\\n=6\end{matrix}\right.\)
Vậy x∈{0,2,-4,6}
2. Ta có:
x-y-z=0 ⇒\(\left\{{}\begin{matrix}x=y+z\\y=x-z\\z=x-y\end{matrix}\right.\)
Thay vào biểu thức ta được:
\(B=\left(1-\frac{x-y}{x}\right)\left(1-\frac{y+z}{y}\right)\left(1+\frac{x-z}{z}\right)\)
⇒\(B=\frac{x-x+y}{x}.\frac{y-y-z}{y}.\frac{z+x-z}{z}\)
⇒\(B=\frac{y.\left(-z\right).x}{x.y.z}=\frac{\left(-1\right)xyz}{xyz}=-1\)
Vậy biểu thức B có giá trị là -1
Đặt \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=ak\\y=bk\\z=ck\end{matrix}\right.\)
Ta có: \(H=\frac{xyz\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)
\(=\frac{ak\cdot bk\cdot ck\cdot\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc\cdot\left(ak+bk\right)\cdot\left(bk+ck\right)\cdot\left(ck+ak\right)}\)
\(=\frac{k^3\cdot abc\cdot\left(a+b\right)\left(b+c\right)\left(c+a\right)}{k^3\cdot abc\cdot\left(a+b\right)\left(b+c\right)\left(c+a\right)}=1\)
Vậy: H=1
đặt \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=k\Leftrightarrow\left\{{}\begin{matrix}x=ak\\y=bk\\z=ck\end{matrix}\right.\)
theo giả thiết ta có \(H=\frac{xyz\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)
thay \(H=\frac{ak.bk.ck\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc\left(ak+bk\right)\left(bk+ck\right)\left(ck+ak\right)}\)
\(\Leftrightarrow H=\frac{k^3abc\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc\left[k\left(a+b\right)\right]\left[k\left(b+c\right)\right]\left[k\left(c+a\right)\right]}\)
\(\Leftrightarrow H=\frac{k^3abc\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc.k\left(a+b\right).k\left(b+c\right).k\left(c+a\right)}\)
\(\Leftrightarrow H=\frac{k^3abc\left(a+b\right)\left(b+c\right)\left(c+a\right)}{k^3abc\left(a+b\right)\left(b+c\right)\left(c+a\right)}=1\)
Vậy H = 1
b) Ta có:
\(\frac{1}{c}=\frac{1}{2}.\left(\frac{1}{a}+\frac{1}{b}\right)\)
\(\Rightarrow\frac{1}{c}.2=\frac{a}{ab}+\frac{b}{ab}\)
\(\Rightarrow\frac{2}{c}=\frac{a+b}{ab}.\)
\(\Rightarrow2ab=\left(a+b\right).c\)
\(\Rightarrow ab+ab=ac+bc\)
\(\Rightarrow ab-bc=ac-ab\)
\(\Rightarrow b.\left(a-c\right)=a.\left(c-b\right)\)
\(\Rightarrow\frac{a}{b}=\frac{a-c}{c-b}\left(đpcm\right).\)
Chúc bạn học tốt!
1.
\(10x=|x+\dfrac{1}{10}|+|x+\dfrac{2}{10}|+...+|x+\dfrac{9}{10}| \ge 0\)
\(\Rightarrow x\ge0\)
\(pt\Leftrightarrow x+\frac{1}{10}+x+\frac{2}{10}+...+x+\frac{9}{10}=10x\)
\(\Leftrightarrow x=\frac{1}{10}+\frac{2}{10}+...+\frac{9}{10}=\frac{9}{2}\)
\(\Rightarrow x=\frac{9}{2}\)
4.
Áp dụng tính chất dãy tỉ số bằng nhau
\(\frac{a}{b+3c}=\frac{b}{c+3a}=\frac{c}{a+3b}=\frac{a+b+c}{4\left(a+b+c\right)}=\frac{1}{4}\)
\(\Rightarrow\left\{{}\begin{matrix}4a=b+3c\left(1\right)\\4b=c+3a\left(2\right)\\4c=a+3b\left(3\right)\end{matrix}\right.\)
Từ \(\left(1\right);\left(2\right)\Rightarrow4a=b+3\left(4b-3a\right)\)
\(\Rightarrow12a=12b\Rightarrow a=b\left(4\right)\)
Từ \(\left(1\right);\left(3\right)\Rightarrow4c=a+3\left(4a-3c\right)\)
\(\Rightarrow12a=12c\Rightarrow a=c\left(5\right)\)
Từ \(\left(4\right);\left(5\right)\Rightarrow a=b=c\left(đpcm\right)\)
a) \(A=x\cdot\left(-1\right)^n\cdot\left|x\right|\)
\(A=x\cdot\left(-1\right)\cdot x\)
\(A=-x^2\)
b) \(\frac{x}{y}-\frac{2}{3}=\frac{y}{z}-\frac{4}{5}=\frac{z}{t}-\frac{6}{7}=0\)và \(x+y+z+t=315\)
Xét :
\(\frac{x}{y}-\frac{2}{3}=0\Leftrightarrow\frac{x}{y}=\frac{2}{3}\Leftrightarrow\frac{x}{2}=\frac{y}{3}\Leftrightarrow\frac{x}{8}=\frac{y}{12}\)
\(\frac{y}{z}-\frac{4}{5}=0\Leftrightarrow\frac{y}{z}=\frac{4}{5}\Leftrightarrow\frac{y}{4}=\frac{z}{5}\Leftrightarrow\frac{y}{12}=\frac{z}{15}\)
\(\frac{z}{t}-\frac{6}{7}=0\Leftrightarrow\frac{z}{t}=\frac{6}{7}\Leftrightarrow\frac{z}{6}=\frac{t}{7}\Leftrightarrow\frac{z}{15}=\frac{t}{\frac{35}{2}}\)
\(\Rightarrow\frac{x}{8}=\frac{y}{12}=\frac{z}{15}=\frac{t}{\frac{35}{2}}\) và \(x+y+z+t=315\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{x}{8}=\frac{y}{12}=\frac{z}{15}=\frac{t}{\frac{35}{2}}=\frac{x+y+z+t}{8+12+15+\frac{35}{2}}=\frac{315}{\frac{105}{2}}=6\)
\(\frac{x}{8}=6\Leftrightarrow x=48\)
\(\frac{y}{12}=6\Leftrightarrow y=72\)
\(\frac{z}{15}=6\Leftrightarrow z=90\)
\(\frac{t}{\frac{35}{2}}=6\Leftrightarrow t=105\)
ta có
\(\frac{x}{y}-\frac{2}{3}=0\Leftrightarrow\frac{x}{y}=\frac{2}{3}\Leftrightarrow\frac{x}{2}=\frac{y}{3}\)
\(\frac{y}{z}-\frac{4}{5}=0\Leftrightarrow\frac{y}{z}=\frac{4}{5}\Leftrightarrow\frac{y}{4}=\frac{z}{5}\)
\(\frac{z}{t}-\frac{6}{7}=0\Leftrightarrow\frac{z}{t}=\frac{6}{7}\Leftrightarrow\frac{z}{7}=\frac{t}{6}\)
ta lại có
\(\hept{\begin{cases}\frac{x}{2}=\frac{y}{3}\\\frac{y}{4}=\frac{z}{5}\end{cases}\Leftrightarrow\hept{\begin{cases}\frac{x}{8}=\frac{y}{12}\\\frac{y}{12}=\frac{z}{15}\end{cases}}}\Leftrightarrow\frac{x}{8}=\frac{y}{12}=\frac{z}{15}\left(1\right)\)
\(\hept{\begin{cases}\frac{y}{12}=\frac{z}{15}\\\frac{z}{7}=\frac{t}{6}\end{cases}\Leftrightarrow\hept{\begin{cases}\frac{y}{84}=\frac{z}{105}\\\frac{z}{105}=\frac{t}{90}\end{cases}}}\Leftrightarrow\frac{y}{84}=\frac{z}{105}=\frac{t}{90}\left(2\right)\)
ta kết hợp (1) và (2)
\(\hept{\begin{cases}\frac{x}{8}=\frac{y}{12}=\frac{z}{15}\\\frac{y}{84}=\frac{z}{105}=\frac{t}{90}\end{cases}}\Leftrightarrow\frac{x}{57}=\frac{y}{84}=\frac{z}{105}=\frac{t}{90}\)và \(x+y+z+t=315\)
theo tính chất dãy tỉ số = nhau
có \(\frac{x}{57}=\frac{y}{84}=\frac{z}{105}=\frac{t}{90}=\frac{x+y+z+t}{57+84+105+90}=\frac{315}{336}=\frac{15}{16}\)
thay vào
Bài 2:
c) \(3x-\left|2x+1\right|=2\)
\(\Rightarrow\left|2x+1\right|=3x-2\)
\(\Rightarrow\left[{}\begin{matrix}2x+1=3x-2\\2x+1=2-3x\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x-3x=\left(-2\right)-1\\2x+3x=2-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}-1x=-3\\5x=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3:1\\x=1:5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=\frac{1}{5}\end{matrix}\right.\)
Vậy \(x\in\left\{3;\frac{1}{5}\right\}.\)
Chúc bạn học tốt!
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