a) x2 + x = 0
=> x( x+ 1 ) = 0
=> x = 0
hoặc x = -1
b) b, (x-1)x+2 = (x-1)x+4
=> x + 2 = x + 4
=> 0x = 2 ( ktm)
Vậy ko có giá trị x nào thoả mãn đk
d) Ta có: x-1/x+5 = 6/7
=>(x-1).7 = (x+5).6
=>7x-7 = 6x+ 30
=> 7x-6x = 7+30
=> x = 37
Vậy x = 37
e, x2/ 6= 24/25
=> x2 . 25 = 6 . 24
⇒
Vậy
g, x-2/ x-1= x+4/ x+7
Trả lời:
\(a,x^2+x=0\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}}\)
Vậy x = 0; x = - 1
\(b,\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\)
\(\Leftrightarrow\left(x-1\right)^{x+2}-\left(x-1\right)^{x+4}=0\)
\(\Leftrightarrow\left(x-1\right)^{x+2}.\left[1-\left(x-1\right)^2\right]=0\)
\(\Leftrightarrow\left(x-1\right)^{x+2}.\left(1-x^2+2x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)^{x+2}\left(2x-x^2\right)=0\)
\(\Leftrightarrow\left(x-1\right)^{x+2}.x\left(2-x\right)=0\)
+) TH1: ( x - 1 )x+2 = 0 <=> x - 1 = 0 <=> x = 1
+) TH2: x = 0
+) TH3: 2 - x = 0 <=> x = 2
Vậy x = 1; x = 0; x = 2
\(c,\frac{1}{4}\cdot\frac{2}{6}\cdot\frac{3}{8}\cdot\frac{4}{10}\cdot\frac{5}{12}\cdot...\cdot\frac{30}{62}\cdot\frac{31}{64}\)\(=2^x\)
\(\Leftrightarrow\frac{1\cdot2\cdot3\cdot4\cdot5\cdot...\cdot30\cdot31}{4\cdot6\cdot8\cdot10\cdot12\cdot...\cdot62\cdot64}=2^x\)
\(\Leftrightarrow\frac{1\cdot2\cdot3\cdot4\cdot5\cdot...\cdot30\cdot31}{2\cdot2\cdot2\cdot3\cdot2\cdot4\cdot2\cdot5\cdot2\cdot6\cdot...\cdot2\cdot31\cdot2\cdot32}=2^x\)
\(\Leftrightarrow\frac{1\cdot2\cdot3\cdot4\cdot5\cdot...\cdot31\cdot32}{\left(2\cdot2\cdot2\cdot2\cdot2\cdot...\cdot2\cdot2\right)\cdot\left(2\cdot3\cdot4\cdot5\cdot6\cdot...\cdot31\cdot32\right)}=2^x\)
\(\Leftrightarrow\frac{1}{2^{31}}=2^x\)
\(\Leftrightarrow2^{31}.2^x=1\)
\(\Leftrightarrow2^{31+x}=2^0\)
\(\Leftrightarrow31+x=0\)
\(\Leftrightarrow x=-31\)
Vậy x = - 31
\(d,\frac{x-1}{x+5}=\frac{6}{7}\)
\(\Leftrightarrow7\left(x-1\right)=6\left(x+5\right)\)
\(\Leftrightarrow7x-7=6x+30\)
\(\Leftrightarrow7x-6x=30+7\)
\(\Leftrightarrow x=37\)
Vậy x = 37
\(e,\frac{x^2}{6}=\frac{24}{25}\)
\(\Leftrightarrow25x^2=6.24\)
\(\Leftrightarrow\left(5x\right)^2=144\)
\(\Leftrightarrow\left(5x\right)^2=\left(\pm12\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}5x=12\\5x=-12\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{12}{5}\\x=-\frac{12}{5}\end{cases}}}\)
Vậy x = 12/5; x = - 12/5
\(g,\frac{x-2}{x-1}=\frac{x+4}{x+7}\) \(\left(ĐKXĐ:x\ne1;x\ne-7\right)\)
\(\Leftrightarrow\left(x-2\right)\left(x+7\right)=\left(x-1\right)\left(x+4\right)\)
\(\Leftrightarrow x^2+7x-2x-14=x^2+4x-x-4\)
\(\Leftrightarrow x^2+7x-2x-14-x^2-4x+x+4=0\)
\(\Leftrightarrow2x-10=0\)
\(\Leftrightarrow2x=10\)
\(\Leftrightarrow x=5\)
Vậy x = 5
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