a,(3-4x)^2=16(x-3)^2
b,(x^2+x+1)^2=(4x-1)^2
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NN
1
LT
24 tháng 2 2020
a) 3-4x = 4.(x-3) hoặc 3-4x = -4.(x-3)
3-4x=4x-12 hoặc 3-4x = -4x +12
8x=15 hoặc -4x+4x=12-3
x=15/8
b) x^2+x+1=4x-1 hoặc x^2+x+1= -(4x-1)
x^2-3x+2=0 hoặc x^2+5x=0
TH1: x^2-3x+2=0
x^2-x-2x+2=0
(x^2-x)-(2x-2)=0
x(x-1)-2(x-1)=0
(x-1).(x-2)=0
x=1 hoặc x=2
TH2: x^2+5x=0
x.(x+5)=0
x=0 hoặc x=-5
Các bạn tự đáp số nhé
D
27 tháng 6 2018
\(a,\)
\(A=\left(\frac{4x}{x+2}-\frac{x^3-8}{x^3+8}.\frac{4x^2-4x+16}{x^2-4}\right):\frac{16}{x+2}.\frac{x^2+3x+2}{x^2+x+1}\)\(ĐKXĐ:x\ne\pm2\)
\(A=[\frac{4x}{x+2}-\frac{\left(x-2\right)\left(x^2+2x+4\right).4\left(x^2-2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)\left(x-2\right)\left(x+2\right)}]:\frac{16}{x+2}.\frac{\left(x+1\right)\left(x+2\right)}{x^2+x+1}\)
\(A=[\frac{4x}{x+2}-\frac{4\left(x^2+2x+4\right)}{\left(x+2\right)^2}].\frac{x+2}{16}.\frac{\left(x+1\right)\left(x+2\right)}{x^2+x+1}\)
\(A=\frac{4x^2+8x-4x^2-8x-16}{\left(x+2\right)^2}.\frac{x+2}{16}.\frac{\left(x+1\right)\left(x+2\right)}{x^2+x+1}\)
\(A=\frac{16\left(x+2\right)}{\left(x+2\right)^2.16}.\frac{\left(x+1\right)\left(x+2\right)}{x^2+x+1}\)
\(A=\frac{-\left(x+1\right)}{x^2+x+1}\)
\(B=\frac{x^2+x-2}{x^3-1}\)\(ĐKXĐ:x\ne1\)
\(B=\frac{\left(x+2\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(B=\frac{x+2}{x^2+x+1}\)
\(b,\)
Ta có:
\(A+B=\frac{-\left(x+1\right)}{x^2+x+1}+\frac{x+2}{x^2+x+1}\)
\(=\frac{-x-1+x+2}{x^2+x+1}\)
\(=\frac{1}{x^2+x+1}\)
\(\Rightarrow A+B=\frac{1}{x^2+x+1}=\frac{1}{x^2+2.x.\left(\frac{1}{2}\right)^2+\frac{3}{4}}=\frac{1}{\left(x+\frac{1}{2}\right)^2}+\frac{3}{4}\)
Vì:\(\left(x+\frac{1}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\forall x\)
\(\Rightarrow\frac{1}{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}\le\frac{1}{\frac{3}{4}}\)
\(\Rightarrow A+B\le\frac{4}{3}\)
\(\Rightarrow GTLN\)của \(A+B=\frac{4}{3}\Leftrightarrow x+\frac{1}{2}=0\)
\(\Leftrightarrow x=\frac{-1}{2}\left(TMĐK\right)\)
Vậy........
26 tháng 6 2018
Câu d đề có đúng ko bn
mk thấy hơi sai Nguyen Thi tuong Vi
VP
3
KK
6 tháng 9 2021
1. \(\sqrt{x^2-4x+3}=x-2\)
<=> x2 - 4x + 3 = (x - 2)2
<=> x2 - 4x + 3 = x2 - 4x + 4
<=> x2 - x2 - 4x + 4x = 1
<=> 0 = 1 (Vô lí)
vậy PT có nghiệm là S = \(\varnothing\)
KK
6 tháng 9 2021
2. \(\sqrt{4x^2-4x+1}=x-1\)
<=> \(\sqrt{\left(2x-1\right)^2}=x-1\)
<=> 2x - 1 = x - 1
<=> 2x - x = -1 + 1
<=> x = 0
HT
2
28 tháng 6 2017
a) \(\left(x-3\right)\left(x+3\right)-\left(x+1\right)^2\) = \(x^2-9-\left(x^2+2x+1\right)\)
\(x^2-9-x^2-2x-1\) = \(-2x-10\)
b) \(\left(4x-3\right)\left(4x+3\right)-16x^2\) = \(16x^2-9-16x^2=-9\)
c) \(\left(x+4\right)\left(x^2-4x+16\right)-x^3\) = \(x^3-4x^2+16x+4x^2-16x+64-x^3\)
= \(64\)
TN
28 tháng 6 2017
\(a,\left(x-3\right)\left(x+3\right)-\left(x+1\right)^2=x^2-9-x^2-2x-1=-10-2x\) \(b,\left(4x-3\right)\left(4x+3\right)-16x^2=16x^2-9-16x^2=-9\)\(c,\left(x+4\right)\left(x^2-4x+16\right)-x^3=x^3+64-x^3=64\)
25 tháng 8 2019
a) \(\left(x-3\right)^2-4=0\)
\(\left(x-3\right)^2=0+4\)
\(\left(x-3\right)^2=4\)
\(\left(x-3\right)^2=\pm4\)
\(\left(x-3\right)^2=\pm2^2\)
\(\orbr{\begin{cases}x-3=2\\x-3=-2\end{cases}}\)
\(\orbr{\begin{cases}x=5\\x=1\end{cases}}\)
b) \(\left(2x+3\right)^2-\left(2x+1\right)\left(2x-1\right)=22\)
\(4x^2+12x+9-4x^2+1=22\)
\(12x+10=22\)
\(12x=22-10\)
\(12x=12\)
\(x=1\)
c) \(\left(4x+3\right)\left(4x-3\right)-\left(4x-5\right)^2=16\)
\(16x^2-9-16x^2+40x-25=16\)
\(-34+40x=16\)
\(40x=16+34\)
\(40x=50\)
\(x=\frac{50}{40}=\frac{5}{4}\)
d) \(x^3-9x^2+27x-27=-8\)
\(x^3-9x^2+27x-27+8=0\)
\(x^3-9x^2+27x-19=0\)
\(\left(x^2-8x+19\right)\left(x-1\right)=0\)
Vì \(\left(x^2-8x+19\right)>0\) nên:
\(x-1=0\)
\(x=1\)
e) \(\left(x+1\right)^3-x^2\left(x+3\right)=2\)
\(x^3+2x^2+x+x^2+2x+1-x^2-3x^2=2\)
\(3x+1=2\)
\(3x=2-1\)
\(3x=1\)
\(x=\frac{1}{3}\)
NM
25 tháng 8 2019
b) ( 2x+3)^2 - (2x+1)(2x-1) =22
=> 4x2+12x+9-4x2+1=22
=> 12x=12
=>x=1
c) (4x+3)(4x-3) -(4x-5)^2 =16
=>16x2-9-16x2+40x-25=16
=>40x=50
=>x=4/5
NB
25 tháng 8 2019
a)\(\left(x-13\right)^2-4=0\\\left(x-13\right)^2=4\\ \left(x-13\right)^2=2^2\\ \Rightarrow\left\{{}\begin{matrix}x-13=2\\x-13=-2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}15\\-11\end{matrix}\right.\)
vậy...
Lời giải:
a)
$(3-4x)^2=16(x-3)^2=4^2(x-3)^2=(4x-12)^2$
$\Leftrightarrow [(3-4x)-(4x-12)][(3-4x)+(4x-12)]=0$
$\Leftrightarrow (15-8x)(-9)=0$
$\Rightarrow 15-8x=0\Rightarrow x=\frac{15}{8}$
b)
$(x^2+x+1)^2=(4x-1)^2$
$\Leftrightarrow (x^2+x+1)^2-(4x-1)^2=0$
$\Leftrightarrow (x^2+x+1-4x+1)(x^2+x+1+4x-1)=0$
$\Leftrightarrow (x^2-3x+2)(x^2+5x)=0$
$\Leftrightarrow (x-1)(x-2)x(x+5)=0$
\(\Rightarrow \left[\begin{matrix} x-1=0\\ x-2=0\\ x=0\\ x+5=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=1\\ x=2\\ x=0\\ x=-5\end{matrix}\right.\)
a) 9 - 24x + 16x2 = 16(x2 - 6x + 9)
=> 16x2 - 24x + 9 = 16x2 - 96x + 144
=> -24x + 96x = 144 - 9
=> 72x = 135
=> x = \(\frac{15}{8}\)
b) (x2 + x + 1)2 = (4x - 1)2
=> x4 + x2 + 1 + 2x3 + 2x + 2x2 = 16x2 - 8x + 1
=> x4 + 2x3 + 3x2 + 2x + 1 = 16x2 - 8x + 1
=> x4 + 2x3 - 13x2 + 10x = 0
=> x4 - x3 + 3x3 - 3x2 - 10x2 + 10x = 0
=> (x3 + 3x2 - 10x)(x - 10) = 0
=> x(x2 + 3x - 10)(x - 10) = 0
=> x(x - 2)(x+5)(x-10) = 0
=> \(\left[{}\begin{matrix}x=0\\x=2\\x=-5\\x=10\end{matrix}\right.\)
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