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Câu 1:Ta có:
a) \(\left|x-3\right|=5\Leftrightarrow\left[{}\begin{matrix}x-3=5\\x-3=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)
b) \(\left|2x+3\right|=2.\left|4-x\right|\)
+)Xét \(\left\{{}\begin{matrix}2x+3\ge0\\4-x\ge0\end{matrix}\right.\) \(\Leftrightarrow\dfrac{-3}{2}\le x\le4\)
Khi đó \(2x+3=2\left(4-x\right)\Leftrightarrow2x+3=8-2x\Leftrightarrow4x=5\Leftrightarrow x=\dfrac{5}{4}\left(tm\right)\)
+) Xét \(\left\{{}\begin{matrix}2x+3\ge0\\4-x\le0\end{matrix}\right.\) \(\Leftrightarrow x\ge4\)
Khi đó: \(2x+3=2\left(x-4\right)=2x-8\Leftrightarrow0x=-11\left(vl\right)\)
+) Xét \(\left\{{}\begin{matrix}2x+3\le0\\4-x\ge0\end{matrix}\right.\) \(\Leftrightarrow x\le\dfrac{-3}{2}\)
Khi đó: \(-\left(2x+3\right)=2.\left(4-x\right)\Leftrightarrow-2x-3=8-2x\left(vl\right)\)
+)Xét \(\left\{{}\begin{matrix}2x+3\le0\\4-x\le0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{-3}{2}\\x\ge4\end{matrix}\right.\) \(\left(vl\right)\)
Vậy...
c) ĐKXĐ : \(3-x\ge0\Leftrightarrow x\le3\)
+)Xét \(x^{^2}-3x+1\ge0\)
\(\Leftrightarrow x^2-3x+1=3-x\Leftrightarrow x^2-2x-2=0\)
\(\Leftrightarrow x^2-2x+1=3\Leftrightarrow\left(x-1\right)^2=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=\sqrt{3}\\x-1=-\sqrt{3}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=1+\sqrt{3}\left(tm\right)\\x=1-\sqrt{3}\left(tm\right)\end{matrix}\right.\)
+)Xét \(x^{^2}-3x+1\le0\)
\(\Leftrightarrow-\left(x^2-3x+1\right)=3-x\)
\(\Leftrightarrow x^2-3x+1=x-3\Leftrightarrow x^2-4x+4=0\)
\(\Leftrightarrow\left(x-2\right)^2=0\Leftrightarrow x-2=0\Leftrightarrow x=2\left(tm\right)\)
Vậy...
Câu 2:
Ta có:
Phương trình \(\left(x+3\right)\left(x^2-2x+m-1\right)=0\) có một nghiệm là \(x=-3\)
\(\Rightarrow\)Phương trình \(\left(x+3\right)\left(x^2-2x+m-1\right)=0\) có ba nghiệm phân biệt khi và chỉ khi \(x^2-2x+m-1=0\) có 2 nghiệm phân biệt và khác \(-3\)
Ta có: \(x^2-2x+m-1=0\) có 2 nghiệm phân biệt khi và chỉ khi \(\text{△}>0\Leftrightarrow8-4m>0\Leftrightarrow m< 2\)
Gọi \(x_1\) và \(x_2\) là 2 nghiệm của phương trình \(x^2-2x+m-1=0\).Theo hệ thức Vi-ét ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{-2}{1}=2\\x_1x_2=\dfrac{m-1}{1}=m-1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x_1=2-x_2\\\left(2-x_2\right).x_2=m-1\end{matrix}\right.\)
Nếu \(x_2\ne-3\) thì \(m-1\ne-15\Leftrightarrow m\ne-14\).
Do vai trò của \(x_1\) và \(x_2\) là như nhau nên \(x^2-2x+m-1=0\) có 2 nghiệm phân biệt và khác \(-3\) khi và chỉ khi: \(\left\{{}\begin{matrix}m< 2\\m\ne-14\end{matrix}\right.\)
Bài 1:
a: \(\Leftrightarrow x^2-5x+6< =0\)
=>(x-2)(x-3)<=0
=>2<=x<=3
b: \(\Leftrightarrow\left(x-6\right)^2< =0\)
=>x=6
c: \(\Leftrightarrow x^2-2x+1>=0\)
\(\Leftrightarrow\left(x-1\right)^2>=0\)
hay \(x\in R\)
\(a,A=\left\{0;1;2;3;4\right\}\\ b,B=\left\{-16;-13;-10;-7;-4;-1;2;5;8\right\}\\ c,C=\left\{-9;-8;-7;...;7;8;9\right\}\\ d,x^2-3x+1=0\\ \Delta=9-4=5\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3-\sqrt{5}}{2}\\x=\dfrac{3+\sqrt{5}}{2}\end{matrix}\right.\\ \Leftrightarrow D=\left\{\dfrac{3-\sqrt{5}}{2};\dfrac{3+\sqrt{5}}{2}\right\}\)
\(e,2x^3-5x^2+2x=0\\ \Leftrightarrow x\left(x-2\right)\left(2x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=\dfrac{1}{2}\left(ktm\right)\end{matrix}\right.\\ \Leftrightarrow E=\left\{0;2\right\}\\ f,F=\left\{0;3;6;9;12;15;18\right\}\)
a: \(A=\left\{0;1;2;3;4;5\right\}\)
b: \(B=\left\{2;3;4;5\right\}\)
c: \(C=\left\{0;1;-1;2;-2;3;-3\right\}\)
1, a,\(\left(-7x^2\right)\left(3x^2-x-2\right)\)
\(=-21x^4+7x^3+14x^2\)
\(b,\left(2x^3-3x^2-10x+3\right):\left(x-3\right)\)
2,\(a,\left(x-3\right)\left(x^2+1\right)-\left(x-3\right)\left(x^2+3x+9\right)\)
\(=x^3+x-3x^2-3-x^3+27\)
\(=-3x^2+x+24\)
\(b,\left(2x+1\right)^2+\left(2x-1\right)^2+2\left(4x^2-1\right)\)
\(=4x^2+4x+1+4x^2-4x+1+8x^2-2\)
\(=24x^2\)
\(3,a,x^3-x^2-x+1\)
\(=x^2\left(x-1\right)-\left(x-1\right)\)
\(=\left(x-1\right)^2\left(x+1\right)\)
\(b,3x^2-7x-10\)
\(=3x^2+3x-10x-10\)
\(=3x\left(x+1\right)-10\left(x+1\right)\)
\(=\left(x+1\right)\left(3x-10\right)\)
4, a. Bn kiểm tra lại đề bài nhé
b,\(4x^2-12xy+10y^2\)
\(=\left(4x^2-12xy+9y^2\right)+y^2\)
\(=\left(2x-3y\right)^2+y^2\ge0\forall x,y\)
Bài 2:
a: (x+1)(3-x)=0
=>x+1=0 hoặc 3-x=0
=>x=-1 hoặc x=3
b: (x-2)(2x-1)=0
=>x-2=0 hoặc 2x-1=0
=>x=2 hoặc x=1/2
c: (3x+9)(1-3x)=0
=>1-3x=0 hoặc 3x+9=0
=>x=1/3 hoặc x=-3
d: (x2+1)(81-x2)=0
=>(9+x)(9-x)=0
=>x=-9 hoặc x=9
Bài 1
d, \(x^2+2xy+y^2-2x-2y+1\)
\(\Rightarrow x^2+y^2=1+2xy-2y-2x\)
\(\Rightarrow\left(x+y-1\right)^2\)
Bài 2:
a, \(\left(x+1\right)\left(x+1\right)=\left(x+2\right)\left(x+5\right)\)
\(\Leftrightarrow\left(x+1\right)^2=x^2+5x+2x+10\)
\(\Leftrightarrow x^2+2x+1=x^2=5x+2x+10\)
\(\Leftrightarrow-5x=9\)
\(\Leftrightarrow x=-\frac{9}{5}\)
b,\(\left(x+3\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-5\end{matrix}\right.\)
c, \(4x^2-9=0\)
\(\Leftrightarrow4x^2=9\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{3}{2}\\\frac{3}{2}\end{matrix}\right.\)
d,\(\left(4x-5\right)^2-\left(3x-4\right)^2=0\)
\(\Leftrightarrow16x^2-40x+25-\left(9x^2-24x+16\right)=0\)
\(\Leftrightarrow16x^2-40x+25-9x^2+24x-16=0\)
\(\Leftrightarrow7x^2-16x+9=0\)
\(\Leftrightarrow x=\frac{-\left(-16\right)\pm\sqrt{\left(-16\right)^2-4.7.9}}{14}\)
\(\Leftrightarrow x=\frac{16\pm\sqrt{256-252}}{14}\)
\(\Leftrightarrow x=\frac{16\pm\sqrt{4}}{14}\)
\(\Leftrightarrow x=\frac{16\pm2}{14}\)
\(\Leftrightarrow x=\left[{}\begin{matrix}\frac{16+2}{14}\\\frac{16-2}{14}\end{matrix}\right.\)
\(\Leftrightarrow x=\left[{}\begin{matrix}\frac{9}{7}\\1\end{matrix}\right.\)
1.a)\(3x-3y+x^2-2xy+y^2\)
\(=3\left(x-y\right)+\left(x-y\right)^2\)
\(=\left(x-y\right)\left(3+x-y\right)\)
d)\(x^2+2xy+y^2-2x-2y+1\)
\(=\left(x+y\right)^2-2\left(x+y\right)+1\)
\(=\left(x+y+1\right)^2\)
2.a)\(\left(x+1\right)\left(x+1\right)=\left(x+2\right)\left(x+5\right)\)
\(\Leftrightarrow\left(x+1\right)^2=x^2+5x+2x+10\)
\(\Leftrightarrow x^2+2x+1-x^2-7x-10=0\)
\(\Leftrightarrow-5x-9=0\)
\(\Leftrightarrow-5x=9\)
\(\Leftrightarrow x=-\frac{9}{5}\). Vậy \(S=\left\{-\frac{9}{5}\right\}\)
b)\(\left(x+3\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-5\end{matrix}\right.\).Vậy \(S=\left\{-3;-5\right\}\)
c)\(4x^2-9=0\)
\(\Leftrightarrow\left(2x+3\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\frac{3}{2}\\x=\frac{3}{2}\end{matrix}\right.\). Vậy \(S=\left\{\pm\frac{3}{2}\right\}\)
d)\(\left(4x-5\right)^2-\left(3x-4\right)^2=0\)
\(\Leftrightarrow\left(4x-5+3x-4\right)\left(4x-5-3x+4\right)=0\)
\(\Leftrightarrow\left(7x-9\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}7x-9=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{9}{7}\\x=1\end{matrix}\right.\). Vậy \(S=\left\{1;\frac{9}{7}\right\}\)
3.Ta có:
Để \(A\left(x\right)⋮B\left(x\right)\) thì: \(m+21⋮2x-3\)
\(\Rightarrow m+21=0\)
\(\Rightarrow m=-21\)
Vậy...!