1. Tìm x
a, | x+3 | + | x+1 | = 3x
b, | 2x-3 | - x = | 2-x |
nói cách làm nữa nha
Tìm x biết
a, 5 . | 2x +3 | = 1/2
b, 1/2 . ( 3x - 5 + | x - 1 | ) = 2x -5
nói cách làm nữa nhé
Tìm x
b, 1/2 . ( 3x - 5 + | x - 1 | ) = 2x -5
nói cách làm nữa nhé
bài 2: Tìm x,biết :(cách làm nữa nhé )
a) 9x2 -49=0
b)(x+3)(x2 -3x+9)-x(x-1)(x+1)-27=0
c) (x-1)(x+2)-x-2=0
d)x(3x +2)+( x+1)2 -(2x-5)(2x+5) = 0
e) (4x+1)(x-2)-(2x-3)(2x+1)=7
a ) \(9x^2-49=9\)
\(\Leftrightarrow9x^2=58\)
\(\Leftrightarrow x^2=29\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=29\\x=-29\end{array}\right.\)
Vậy ......................
b ) \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-1\right)\left(x+1\right)-27=0\)
\(\Leftrightarrow\left(x^3+3^3\right)-x.\left(x^2-1^2\right)-27=0\)
\(\Leftrightarrow x^3+27-x^3+x-27=0\)
\(\Leftrightarrow x=0\)
c ) \(\left(x-1\right)\left(x+2\right)-x-2=0\)
\(\Leftrightarrow x^2+2x-x-2-x-2=0\)
\(\Leftrightarrow x^2-4=0\)
\(\Leftrightarrow x^2=4\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=-2\end{array}\right.\)
Vây .....................
d với e thì tách hết ra, tự triệt tiêu là ra kết quả, dễ mà :) @La Thị Thu Phượng
a) \(9x^2-49=0\)
\(\left(3x\right)^2-7^2=0\)
\(\Rightarrow\left(3x+7\right)\left(3x-7\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}3x+7=0\\3x-7=0\end{array}\right.\Rightarrow\left[\begin{array}{nghiempt}x=-\frac{7}{3}\\x=\frac{7}{3}\end{array}\right.\)
b) \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-1\right)\left(x+1\right)-27=0\)
\(x^3+27-x\left(x^2-1\right)-27=0\)
\(x^3+27-x^3+x-27=0\)
\(x=0\)
c) \(\left(x-1\right)\left(x+2\right)-x-2=0\)
\(x^2+2x-x-2-x-2=0\)
\(x^2+4=0\)
\(x^2=-4\) (loại vì \(x^2\ge0\) với mọi x)
d) \(x\left(3x+2\right)+\left(x+1\right)^2-\left(2x-5\right)\left(2x+5\right)=0\)
\(3x^2+2x+x^2+2x+1-4x^2+25=0\)
\(4x+26=0\)
\(4x=-26\)
\(x=-\frac{13}{2}\)
e) \(\left(4x+1\right)\left(x-2\right)-\left(2x-3\right)\left(2x+1\right)=7\)
\(4x^2-8x+x-2-4x^2-2x+6x+3=7\)
\(-3x+1=7\)
\(-3x=6\)
\(x=-2\)
Tìm giá trị Nhỏ Nhất
a, \(A=9x^2-2x+15\)
b, \(B=3x^2+x+1\)
c,\(C=x^2-6y+4x+y^2+38\)
nói cách làm nữa nha
a) \(A=9x^2-2x+15\)
\(A=9x^2-2x+\frac{1}{9}+\frac{134}{9}\)
\(A=\left(3x+\frac{1}{3}\right)^2+\frac{134}{9}\)
Có: \(\left(3x+\frac{1}{3}\right)^2\ge0\Rightarrow\left(3x+\frac{1}{3}\right)^2+\frac{134}{9}\ge\frac{134}{9}\)
Dấu '=' xảy ra khi: \(\left(3x+\frac{1}{3}\right)^2=0\Rightarrow3x+\frac{1}{3}=0\Rightarrow x=-\frac{1}{9}\)
Vậy: \(Min_A=\frac{134}{9}\) tại \(x=-\frac{1}{9}\)
b) \(B=3x^2+x+1\)
\(B=3x^2+x+\frac{1}{12}+\frac{11}{12}\)
\(B=\left(\sqrt{3}x+\sqrt{\frac{1}{12}}\right)^2+\frac{11}{12}\)
Có: \(\left(\sqrt{3}x+\sqrt{\frac{1}{12}}\right)^2\ge0\Rightarrow\left(\sqrt{3}x+\sqrt{\frac{1}{12}}\right)^2+\frac{11}{12}\ge\frac{11}{12}\)
Dấu '=' xảy ra khi: \(\left(\sqrt{3}x+\sqrt{\frac{1}{12}}\right)^2=0\Rightarrow\sqrt{3}x+\sqrt{\frac{1}{12}}=0\Rightarrow x=-\frac{1}{6}\)
Vậy: \(Min_B=\frac{11}{12}\) tại \(x=-\frac{1}{6}\)
c) \(C=x^2-6y+4x+y^2+38\)
\(C=\left(x^2+4x+4\right)+\left(y^2-6y+9\right)+25\)
\(C=\left(x+2\right)^2+\left(y-3\right)^2+25\)
Có: \(\left(x+2\right)^2+\left(y-3\right)^2\ge0\Rightarrow\left(x+2\right)^2+\left(y-3\right)^2+25\ge25\)
Dấu = xảy ra khi: \(\hept{\begin{cases}\left(x+2\right)^2=0\\\left(y-3\right)^2=0\end{cases}}\Rightarrow\hept{\begin{cases}x+2=0\\y-3=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-2\\y=3\end{cases}}\)
Vậy: \(Min_C=25\) tại \(\hept{\begin{cases}x=-2\\y=3\end{cases}}\)
Tìm x thuộc Z để min
a, \(\frac{1}{x-3}\)
b, \(\frac{7-x}{x-6}\)
c, \(\frac{3x-19}{x}\)
d, \(\frac{8x+1}{2x-2}\)
nói cách làm nữa nha
tìm x nguyên để giá trị của f(x) chia hết cho giá trị của g(x) biết
a) f(x)=2x3+3x2-x+4; g(x)=2x+1
b) f(x)=3x3-x2+6x;g(x)=3x-1
cách làm nha
Tìm x
c, \(\frac{3x-5}{2x+1}=\frac{6x-7}{4x-3}\)
d,\(\frac{x-1}{4}=\frac{9}{x-1}\)
nói cách làm nữa nhé
c) Bạn dùng "tích trung tỉ bằng tích ngoại tỉ"
d) \(\frac{x-1}{4}=\frac{9}{x-1}\)
=> (x - 1)2 = 4 . 9 = 36 = (+ 6)2
=> x - 1 = 6 hoặc x - 1 = -6
=> x = 7 hoặc x = -5
Tìm x
a) (x + 3)2 + (x + 2)(5 – x) = 1
b/ (2x – 1)2 – ( x – 5)( 4x + 3) = 3
c/ 3x (x – 2) + 4x – 8 = 0
d/ 2x (3x + 5) – 18x – 30 = 0
\(a,\Leftrightarrow x^2+6x+9-x^2+3x+10=1\\ \Leftrightarrow9x=-18\Leftrightarrow x=-2\\ b,\Leftrightarrow4x^2-4x+1-4x^2+17x+15=3\\ \Leftrightarrow13x=-13\Leftrightarrow x=-1\\ c,\Leftrightarrow3x\left(x-2\right)+4\left(x-2\right)=0\\ \Leftrightarrow\left(3x+4\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=2\end{matrix}\right.\\ d,\Leftrightarrow2x\left(3x+5\right)-6\left(3x+5\right)=0\\ \Leftrightarrow\left(x-3\right)\left(3x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{3}\end{matrix}\right.\)
tìm x
a,(x+5)(x-5)-x(x+3)=10 b,(2x+3)(2x-3)-4(x+2)^2=5
c,9x(x+5)-(3x+2)(3x-2)=7 d,(x+1)^3-x(x^2+3x-5)=8
a: =>x^2-25-x^2-3x=10
=>-3x=35
=>x=-35/3
b: =>4x^2-9-4(x^2+4x+4)=5
=>4x^2-9-4x^2-16x-16-5=0
=>-16x-30=0
=>x=-15/8
c: =>9x^2+45x-9x^2+4=7
=>45x=3
=>x=1/15
d: =>x^3+3x^2+3x+1-x^3-3x^2+5x=8
=>8x=7
=>x=7/8