Đại số lớp 7 - Hỏi đáp

\(x^2+5x< 0\)

\(x\left(x+5\right)< 0\)

\(\Leftrightarrow x< 0\)

\(\Leftrightarrow x+5>0\Leftrightarrow x>-5\)

\(-5< x< 0\)

\(x\in\left\{-4;-3;-2;-1\right\}\)

\(\Leftrightarrow x>0\)

\(\Leftrightarrow x-5< 0\Leftrightarrow x< 5\)

\(0< x< 5\)

\(x\in\left\{1;2;3;4\right\}\)

Vậy.......

a, \(\dfrac{5}{6}-\left|2-x\right|=\dfrac{1}{3}\Rightarrow\dfrac{5}{6}-\dfrac{1}{3}=\left|2-x\right|\)

<=> \(\dfrac{1}{2}=\left|2-x\right|\) \(\Leftrightarrow\left[{}\begin{matrix}2-x=\dfrac{1}{2}\\2-x=\dfrac{-1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)

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Mấy câu sau tương tự thôi

b) \(2.\left(3x-1\right).\left(2x+5\right)-6.\left(2x-1\right).\left(x+2\right)=1\)

\(\Leftrightarrow\left(6x-2\right).\left(2x+5\right)-\left(12x-6\right).\left(x+2\right)=1\)

\(\Leftrightarrow12x^2+30x-4x-10-\left(12x^2+24x-6x-12\right)=1\)

\(\Leftrightarrow12x^2+26x-10-12x^2-18x +12=1\)

\(\Leftrightarrow8x+2=1\)

\(\Rightarrow x=\dfrac{-1}{8}\)

c) \(\left(3x-1\right).\left(2x+7\right)-\left(x+1\right).\left(6x-5\right)=\left(x+2\right)-\left(x-5\right)\)

\(\Leftrightarrow6x^2+21x-2x-7-\left(6x^2-5x+6x-5\right)=x+2-x+5\)

\(\Leftrightarrow18x-2-7=0\)

\(\Rightarrow x=\dfrac{9}{18}=\dfrac{1}{2}\)

Bài 2:

a, \(P=x^2-2x+5\)

\(P=x^2-x-x+1+4=\left(x-1\right)^2+4\)

Với mọi giá trị của \(x\in R\)ta có:

\(\left(x-1\right)^2\ge0\Rightarrow\left(x-1\right)^2+4\ge4\)

Hay \(P\ge4\) với mọi giá trị của \(x\in R\).

Để \(P=4\) thì \(\left(x-1\right)^2+4=4\)

\(\Rightarrow x=1\)

Vậy........

b, Xem lại đề.

c, \(M=x^2+y^2-x+6y+10\)

\(M=x^2-\dfrac{1}{2}x-\dfrac{1}{2}x+\dfrac{1}{4}+y^2+3y+3y+9+\dfrac{3}{4}\)

\(M=\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2+\dfrac{3}{4}\)

Với mọi giá trị của \(x;y\in R\)ta có:

\(\left(x-\dfrac{1}{2}\right)^2\ge0;\left(y+3\right)^2\ge0\)

\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2\ge0\)

\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

Hay \(M\ge\dfrac{3}{4}\) với mọi giá trị của \(x;y\in R\).

Để \(M=\dfrac{3}{4}\) thì \(\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2+\dfrac{3}{4}=\dfrac{3}{4}\)

\(\Rightarrow\left\{{}\begin{matrix}\left(x-\dfrac{1}{2}\right)^2=0\\\left(y+3\right)^2=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-3\end{matrix}\right.\)

Vậy............

Chúc bạn học tốt!!!

Bài 1:

a, \(x^2-6x+10=x^2-3x-3x+9+1\)

\(=x.\left(x-3\right)-3.\left(x-3\right)+1=\left(x-3\right)^2+1\)

Với mọi giá trị của \(x\in R\) ta có:

\(\left(x-3\right)^2\ge0\Rightarrow\left(x-3\right)^2+1\ge1>0\)

Vậy................... (đpcm)

b, \(4x-x^2-5=-\left(x^2-4x+5\right)\)

\(=-\left(x^2-2x-2x+4+1\right)\)

\(=-\left[x.\left(x-2\right)-2.\left(x-2\right)+1\right]\)

\(=-\left[\left(x-2\right)^2+1\right]\)

Với mọi giá trị của \(x\in R\) ta có:

\(\left(x-2\right)^2\ge0\Rightarrow\left(x-2\right)^2+1\ge1\)

\(\Rightarrow-\left[\left(x-2\right)^2+1\right]\le-1< 0\)

Vậy............... (đpcm)

Chúc bạn học tốt!!!

Ghi rõ đề hơn chút được hk

a/ \(\left(x-\dfrac{1}{2}\right)^2=0\Rightarrow x-\dfrac{1}{2}=0\Rightarrow x=\dfrac{1}{2}\)

Vậy..........

b/ \(\left(x-2\right)^2=1\)

\(\Rightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

Vậy...............

c/ \(\left(2x-1\right)^3=-8\)

\(\Rightarrow\left(2x-1\right)^3=\left(-4\right)^3\)

\(\Rightarrow2x-1=-4\)

\(\Rightarrow x=\dfrac{-4+1}{2}=-\dfrac{3}{2}\)

Vậy...........

d/ \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)

\(\Rightarrow\left(x+\dfrac{1}{2}\right)^2=\left(\pm\dfrac{1}{4}\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{4}\\x+\dfrac{1}{2}=-\dfrac{1}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{4}\\x=-\dfrac{3}{4}\end{matrix}\right.\)

Vậy........

a) \(\left(x-\dfrac{1}{2}\right)^2=0\)

\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2=0^2\)

\(\Leftrightarrow x-\dfrac{1}{2}=0\)

\(\Leftrightarrow x=0+\dfrac{1}{2}\)

\(\Leftrightarrow x=\dfrac{1}{2}\left(TM\right)\)

Vậy \(x=\dfrac{1}{2}\) là giá trị cần tìm

b) \(\left(x-2\right)^2=1\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-2\right)^2=1^2\\\left(x-2\right)^2=\left(-1\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-2=1\\x-2=\left(-1\right)\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=1\end{matrix}\right.\) \(\left(TM\right)\)

Vậy \(x\in\left\{3;1\right\}\)

c) \(\left(2x-1\right)^3=-8\)

\(\Rightarrow\left(2x-1\right)^3=\left(-2\right)^3\)

\(\Rightarrow2x-1=-2\)

\(\Rightarrow2x=-2+1\)

\(\Rightarrow2x=-1\)

\(\Rightarrow x=\dfrac{-1}{2}\left(TM\right)\)

Vậy \(x=\dfrac{-1}{2}\)

d) \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{4}\\\left(x+\dfrac{1}{2}\right)^2=\dfrac{-1}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{4}\\x+\dfrac{1}{2}=-\dfrac{1}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-1}{4}\\x=\dfrac{-3}{4}\end{matrix}\right.\)

Vậy \(x\in\left\{\dfrac{-1}{4};\dfrac{-3}{4}\right\}\) là giá trị cần tìm

\(\left(x-1\right)\left(x+3\right)< 0\)

\(\Rightarrow\left\{{}\begin{matrix}x-1>0\\x+3< 0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x>1\\x< -3\end{matrix}\right.\) \(\Rightarrow PTVN\)

\(\left\{{}\begin{matrix}x-1< 0\\x+3>0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x< 1\\x>-3\end{matrix}\right.\)\(\Rightarrow-3< x< 1\)

Tập nghiệm : -3 < x < 1

Mình nhớ không lầm thì đây thuộc dạng toán lớp 8 hay sao á

\(\left(x-1\right).\left(x+3\right)< 0\)

\(\Rightarrow\left\{{}\begin{matrix}x-1>0\\x+3>0\\x-1< 0\\x+3< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>1\\x>-3\\x< 1\\x< -3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 1\\x< -3\end{matrix}\right.\)

\(\Rightarrow-3< x< 1\)

Vậy \(x\) nằm trong khoảng \(-3< x< 1\)

$3x+x^2=0$

\(\Rightarrow x.\left(3+x\right)=0\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x+3=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\) Vậy \(x=0\) hoặc \(x=-3\)

\(3x+x^2=0\)

\(\Leftrightarrow x.\left(x+3\right)=0\)

\(\Rightarrow x=0\) hoặc \(x=-3\)

chắc là 120 độ bn ạ

Vì \(\left(x^2+2010\right)\left(x-10\right)=0\) (*) do \(x^2+2010\ge2010,\) với mọi x

Khi đó (*) <=> x-10=0 => x=10

Ta suy ra P = \(\left(10^2-1\right)\left(10^2-2\right)...\left(10^2-2015\right)\)= 99.98.97. ... .2.1.0.(-1).(-2). ... .(-1915) = 0

Vậy giá trị của P = 0

\(3^{x-1}+5.3^{x-1}=162\)

\(3^{x-1}\left(1+5\right)=162\)

\(3^{x-1}.6=162\)

\(3^{x-1}=162:6\)

\(3^{x-1}=27\)

\(3^{x-1}=3^3\)

\(\Leftrightarrow x-1=3\)

\(\Leftrightarrow x=4\)

Vậy \(x=4\) là giá trị cần tìm

\(27^n.9^n=9^{27}:81\) \(\Rightarrow\left(27.9\right)^n=9^{27}:9^2\) \(\Rightarrow\left(3^5\right)^n=9^{25}\) \(\Rightarrow\left(3^5\right)^n=\left(3^2\right)^{25}\) \(\Rightarrow3^{5n}=3^{50}\) \(\Rightarrow5n=50\) \(\Rightarrow n=10\) Vậy số tự nhiên n là 10

\(27^n.9^n\) =\(9^{27}:81\)

\(\Leftrightarrow\left(27.9\right)^n=9^{27}.9^2\)

\(\Leftrightarrow243^n=9^{25}\)

\(\Leftrightarrow3^{5n}=3^{50}\)

\(\Rightarrow5n=50\)

\(\Rightarrow n=50:5\)

\(\Rightarrow n=10\)

Vậy số tự nhiên n cần tìm là 10