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BÀI 1:
Ta có: \(VT=\left(7x+1\right)^2-\left(x+7\right)^2\)
\(=\left(7x+1+x+7\right)\left(7x+1-x-7\right)\)
\(=\left(8x+8\right)\left(6x-6\right)\)
\(=8\left(x+1\right).6\left(x-1\right)\)
\(=48\left(x^2-1\right)=VP\) (đpcm)
Bài 2:
\(16x^2-\left(4x-5\right)^2=15\)
\(\Leftrightarrow\)\(16x^2-16x^2+40x-25=15\)
\(\Leftrightarrow\)\(40x=40\)
\(\Leftrightarrow\)\(x=1\)
Vậy...
Bài 3:
\(A=x^2+2x+3=\left(x+1\right)^2+2\ge2\)
Vậy MIN A = 2 khi x = -1
B1 Xét (7x+1)\(^2\)-(x+7)\(^2\)-48(x\(^2\)-1)
=49\(x^2\)+14x+1-x\(^2\)-14x-49-48x\(^2\)+48
=0
Vậy \(\left(7x+1\right)^2-\left(x+7\right)^2=48\left(x^2-1\right)\)
B2 \(16x^2-\left(4x-5\right)^2=15\)
(4x)\(^2\)-(4x-5)\(^2\)-15=0
(4x-4x+5)(4x+4x-5)-15=09x-5)=0
5(8x-5)-15=0
40x-25-15=0
40x-40=0
x =1
câu B3 mình không bik làm
chúc bạn học tốt ~~~
1. a) 7x2 - 5x - 2 = 7x2 - 7x + 2x - 2 = 7x(x - 1) + 2(x - 1) = (x - 1).(7x + 2)
2. 5(2x - 1)2 - 3(2x - 1) = 0
<=> (2x - 1).[5(2x - 1) - 3] = 0
<=> (2x - 1).(10x - 8) = 0
<=> (2x - 1) = 0 hoặc (10x - 8) = 0
<=> x = 1/2 hoặc x = 4/5
3. x2 - 4x + 7 = (x2 - 4x + 4) + 3 = (x - 2)2 + 3
Do: (x - 2)2 > hoặc = 0 (với mọi x)
Nên (x - 2)2 + 3 > hoặc = 3 (với mọi x)
Hay (x - 2)2 + 3 > 0 (với mọi x) => đpcm
\(A=3\left(x-\frac{5}{6}\right)^2+\frac{11}{12}\)
\(B=2\left(x-\frac{3}{4}\right)^2+\frac{23}{8}\)
\(C=\left(x+\frac{3}{2}\right)^2+\frac{11}{4}\)
\(D=\left(x-5\right)^2+\left(3y+1\right)^2+4\)
\(E=\left(4x+1\right)^2+\left(y-2\right)^2+1\)
\(M=-\left(x+\frac{7}{2}\right)^2-\frac{11}{4}\)
\(N=-5\left(x-\frac{3}{5}\right)^2-\frac{41}{5}\)
\(C\) đề sai ví dụ \(x=3\Rightarrow C=2>0\)
\(D=-5\left(x-\frac{7}{10}\right)^2-\frac{131}{20}\)
a, Xem lại đề:
b, \(16x^2-\left(4x-5\right)^2=15\)
\(\Rightarrow16x^2-\left(16x^2-40x+25\right)=15\)
\(\Rightarrow16x^2-16x^2+40x-25=15\)
\(\Rightarrow40x=40\Rightarrow x=1\)
Chúc bạn học tốt!!!
\(a.\:\left(7x+3\right)^2-4\left(x-1\right)\left(x+1\right)=49\\ 49x^2+42x+9-4x^2+4=49\\ 45x^2+42x+13=49\\ x^2+\dfrac{42}{45}x+\dfrac{13}{45}=\dfrac{49}{45}\\ x^2+2.\dfrac{7}{15}x+\left(\dfrac{7}{15}\right)^2=\dfrac{49}{45}-\dfrac{13}{45}+\left(\dfrac{7}{15}\right)^2\\ \left(x+\dfrac{7}{15}\right)^2=\dfrac{229}{225}\\ \Rightarrow\left[{}\begin{matrix}x+\dfrac{7}{15}=\dfrac{229}{225}\\x+\dfrac{7}{15}=-\dfrac{229}{225}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{124}{225}\\x=-\dfrac{334}{225}\end{matrix}\right.\)
a) 4x2 - 2x + 3 - 4x.(x - 5) = 7x - 3
--> 4x2 - 2x + 3 - 4x2 + 20x = 7x - 3
--> 4x2 - 2x - 4x2 + 20x - 7x = -3 - 3
--> 11x = -6
--> x = \(\frac{-6}{11}\)
b) -3x.(x - 5) + 5.(x - 1) + 3x2 = 4x
--> -3x2 + 15x + 5x - 5 + 3x2 = 4x
--> -3x2 + 15x + 5x + 3x2 - 4x = 5
--> 16x = 5
--> x = \(\frac{5}{16}\)
c) 7x.(x - 2) - 5.(x - 1) = 21x2 - 14x2 + 3
--> 7x2 - 14x - 5x + 5 = 7x2 + 3
--> 7x2 - 14x - 5x - 7x2 = -5 + 3
--> -19x = -2
--> x = \(\frac{2}{19}\)
d) 3.(5x - 1) - x.(x - 2) + x2 - 13x = 7
--> 15x - 3 - x2 + 2x + x2 - 13x = 7
--> 15x - x2 + 2x + x2 - 13x = 3 + 7
--> 4x = 10
--> x = \(\frac{5}{2}\)
e) \(\frac{1}{5}\)x.(10x - 15) - 2x.(x - 5) = 12
--> 2x2 - 3x - 2x2 + 10x = 12
--> 7x = 12
--> x = \(\frac{12}{7}\)
~ Học tốt ~
a) 4x2 - 2x + 3 - 4x(x - 5) = 7x - 3
=> 4x2 - 2x + 3 - 4x2 + 20x = 7x - 3
=> 18x + 3 = 7x - 3
=> 18x - 7x = -3 - 3
=> 11x = -6
=> x = -6/11
b) -3x(x - 5) + 5(x - 1) + 3x2 = 4x
=> -3x2 + 15x + 5x - 5 + 3x2 = 4x
=> 20x - 5 = 4x
=> 20x - 4x = 5
=> 16x = 5
=> x = 5/16
\(c,7x\left(x-2\right)-5\left(x-1\right)=21x^2-14x^2+3\)
\(\Leftrightarrow7x^2-14x-5x+5=7x^2+3\)
\(\Leftrightarrow7x^2-7x^2-19x=3-5\)
\(\Leftrightarrow-19x=-2\)
\(\Leftrightarrow x=\frac{2}{19}\)
Theo đầu bài ta có:
\(\left(7x+1\right)^2-\left(x+7\right)^2=48\left(x^2-1\right)\)
\(\Rightarrow\left[\left(7x+1\right)+\left(x+7\right)\right]\left[\left(7x+1\right)-\left(x+7\right)\right]=\left(7^2-1^2\right)\left(x^2-1^2\right)\)
\(\Rightarrow\left(8x+8\right)\left(6x-6\right)=\left[\left(7+1\right)\left(7-1\right)\right]\left[\left(x+1\right)\left(x-1\right)\right]\)
\(\Rightarrow8\left(x+1\right)\cdot6\left(x-1\right)=8\left(x+1\right)\cdot6\left(x-1\right)\)( đpcm )
Xét \(\left(7x+1\right)^2-\left(x+7\right)^2-48\left(x^2-1\right)\)
\(=49x^2+14x+1-x^2-14x-49-48x^2+48\)
\(=0\)
Vậy \(\left(7x+1\right)^2-\left(x+7\right)^2=48\left(x^2-1\right)\)
Bài 1:
\(16x^2-\left(4x-5\right)^2=15\)
\(\Leftrightarrow\left(4x-4x+5\right)\left(4x+4x-5\right)=15\)
\(\Leftrightarrow5\left(8x-5\right)=15\)
\(\Leftrightarrow8x=8\Leftrightarrow x=1\)
Vậy x = 1
Bài 2:
\(VT=\left(7x+1\right)^2-\left(x+7\right)^2\)
\(=\left(7x+1-x-7\right)\left(7x+1+x+7\right)\)
\(=\left(6x-6\right)\left(8x+8\right)\)
\(=48\left(x-1\right)\left(x+1\right)\)
\(=48\left(x^2-1\right)=VP\)
\(\Rightarrowđpcm\)
thanks