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6) c) x3 - x2 + x = 1
<=> x3 - x2 + x - 1 = 0
<=> (x3 - x2) + (x - 1) = 0
<=> x2 (x - 1) + (x - 1) = 0
<=> (x - 1) (x2 + 1) = 0
=> x - 1 = 0 hoặc x2 + 1 = 0
* x - 1 = 0 => x = 1
* x2 + 1 = 0 => x2 = -1 => x = -1
Vậy x = 1 hoặc x = -1
Bài 5:
a) Đặt \(A=\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^{16}-1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=3^{32}-1\)
\(\Rightarrow A=\frac{3^{32}-1}{8}\)
b) (7x+6)2 + (5-6x)2 - (10-12x)(7x+6)
=(7x+6)2 + (5-6x)2 - 2(5-6x)(7x+6)
\(=\left(7x+6-5+6x\right)^2\)
\(=\left(13x+1\right)^2\)
Bài 1 :
a, \(\left(x-3\right)^2-4=0\Leftrightarrow\left(x-3\right)^2=4\Leftrightarrow\left(x-3\right)^2=\left(\pm2\right)^2\)
TH1 : \(x-3=2\Leftrightarrow x=5\)
TH2 : \(x-3=-2\Leftrightarrow x=1\)
b, \(x^2-2x=24\Leftrightarrow x^2-2x-24=0\)
\(\Leftrightarrow\left(x-6\right)\left(x+4\right)=0\)
TH1 : \(x-6=0\Leftrightarrow x=6\)
TH2 : \(x+4=0\Leftrightarrow x=-4\)
c, \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+2\right)\left(x-2\right)=0\)
\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5\left(x^2-4\right)=0\)
\(\Leftrightarrow2x+30=0\Leftrightarrow x=-15\)
d, tương tự
Ik mk nha, hôm nay ngày mai, ngày kia mk ik 3 lần lại cho bạn (thành 9 lần)
Nhớ kb với mìn lun nha!! Mk rất vui đc làm quen vs bạn, cảm ơn mn nhìu lắm
a) \(A=x^2-8x+17=\left(x-4\right)^2+1\ge1\)
Vậy MIN A = 1 khi x = 4
b) \(T=x^2-4x+7=\left(x-2\right)^2+3\ge3\)
Vậy MIN T = 3 khi x = 2
c) \(H=3x^2+6x-1=3\left(x+1\right)^2-4\ge-4\)
Vậy MIN H = -4 khi x = -1
d) \(E=x^2+y^2-4\left(x+y\right)+16=\left(x-2\right)^2+\left(y-2\right)^2+8\ge8\)
Vậy MIN E = 8 khi x = y = 2
e) \(K=4x^2+y^2-4x-2y+3=\left(2x-1\right)^2+\left(y-1\right)^2+1\ge1\)
Vậy MIN K = 1 khi x = 1/2; y = 1
f) \(M=\frac{3}{2}x^2+x+1=\frac{3}{2}\left(x+\frac{1}{3}\right)^2+\frac{5}{6}\ge\frac{5}{6}\)
Vậy MIN M = 5/6 khi x = -1/3
4.a) \(2x^2-10x-3x-2x^2-26=0\)
\(-13x-26=0\Rightarrow-13\left(x+2\right)=0\)
\(\Rightarrow x=-2\)
b) \(2\left(x+5\right)-x^2-5x=0\)
\(2x+10-x^2-5x=0\Leftrightarrow-x^2-3x+10=0\)
\(-\left(x^2+3x-10\right)=0\)
\(-\left(x^2-2x+5x-10\right)=-\left(x\left(x-2\right)+5\left(x-2\right)\right)=0\)
\(-\left(x-2\right)\left(x+5\right)=0\)
\(\left\{{}\begin{matrix}x-2=0\\x+5=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)
c) \(\left(2x-3\right)^2-\left(x+5\right)^2=0\)
\(\left(2x-3-x-5\right)\left(2x-3+x+5\right)=0\)
\(\left(x-8\right)\left(3x+2\right)=0\)
\(\left\{{}\begin{matrix}x-8=0\\3x+2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=8\\x=-\dfrac{2}{3}\end{matrix}\right.\)
d) \(x^3+x^2-4x-4=0\)
\(x^2\left(x+1\right)-4\left(x+1\right)=0\)
\(\left(x+1\right)\left(x^2-4\right)=\left(x+1\right)\left(x-2\right)\left(x+2\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x+1=0\\x-2=0\\x+2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-1\\x=2\\x=-2\end{matrix}\right.\)
g) \(\left(x-1\right)\left(2x+3-x\right)=0\)
\(\left(x-1\right)\left(x+3\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x-1=0\\x+3=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)
h) \(x^2-4x+8-2x+1=x^2-6x+9=0\)
\(\left(x-3\right)^2=0\Rightarrow x=3\)
bài 5 :
+) ta có : \(A=x^2-4x+18=x^2-4x+4+14\)
\(=\left(x-2\right)^2+14\ge14>0\forall x\Rightarrow\left(đpcm\right)\)
+) ta có : \(B=x^2-x+2=x^2-x+\dfrac{1}{4}+\dfrac{7}{4}\)
\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}>0\forall x\Rightarrow\left(đpcm\right)\)
+) ta có : \(C=x^2+2y^2-2xy-2y+15=x^2-2xy+y^2+y^2-2y+1+14\)
\(=\left(x-y\right)^2+\left(y-1\right)^2+14\ge14>0\forall x\Rightarrow\left(đpcm\right)\)
bài 6 :
+) ta có : \(M=x^2-10x+3=x^2-10x+25-22=\left(x-5\right)^2-22\ge-22\)
\(\Rightarrow M_{min}=-22\) khi \(x=5\)
+) ta có : \(N=x^2+6x-5=x^2+6x+9-14=\left(x+3\right)^2-14\ge-14\)
\(\Rightarrow N_{min}=-14\) khi \(x=-3\)
+) ta có : \(P=x^2+y^2-4x+20=x^2-4x+4+y^2+16=\left(x-2\right)^2+y^2+16\ge16\)
\(\Rightarrow P_{min}=16\) khi \(\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\)
+) ta có : \(Q=x\left(x-3\right)=x^2-3x=x^2-3x+\dfrac{9}{4}-\dfrac{9}{4}=\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{4}\ge\dfrac{-9}{4}\)
\(\Rightarrow Q_{min}=\dfrac{-9}{4}\) khi \(x=\dfrac{3}{2}\)
bài 7 :
+) ta có : \(A=-x^2-12x+3=-\left(x^2+12x+36\right)+39=-\left(x+6\right)^2+39\le39\)
\(\Rightarrow A_{max}=39\) khi \(x=-6\)
+) ta có : \(B=-4x^2+4x+7=-\left(x^2-4x+4\right)+11=-\left(x-2\right)^2+11\le11\)
\(\Rightarrow B_{max}=11\) khi \(x=2\)
bài 8 :
a) ta có : \(16x^2-9=0\Leftrightarrow x^2=\dfrac{9}{16}\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{-3}{4}\end{matrix}\right.\)
b) ta có : \(\left(x-2\right)^2-x^2=4\Leftrightarrow x^2-4x+4-x^2-4=0\Leftrightarrow x=0\)
c) ta có : \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x-7\right)\left(x+7\right)=0\)
\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5x^2+245=0\)
\(\Leftrightarrow2x+255=0\Leftrightarrow x=\dfrac{-255}{2}\)
d) ta có : \(\left(2x-3\right)^2-\left(2x+1\right)\left(2x-1\right)=16\)
\(\Leftrightarrow4x^2-12x+9-4x^2+1-16=0\Leftrightarrow-12x-6=0\Leftrightarrow x=\dfrac{-1}{2}\)
e) ta có : \(\left(x-2\right)\left(x+2\right)-x\left(x-2\right)=1\)
\(\Leftrightarrow x^2-4-x^2+2x-1=0\Leftrightarrow x=\dfrac{5}{2}\)
thank you bạn nha