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 x
a, \(\frac{2x-5}{4}=\frac{x+1}{3}\)
b, \(\frac{x-\frac{1}{2}}{2^3-3^3}=\frac{8x}{16^2-2^7}\)
nói cách làm nữa nhé
Tìm x, biết :
a) \(\left|x-1\right|-2=x-3\)
b) \(\left|2x+3\right|+2x+5=2\)
c ) \(5x-\left|2x+5\right|=3x+5\)
d) \(3x+\left|5-2x\right|=10\)
Mỗi câu trên làm bằng 2 cách nhé !!!
Tìm x
a, \(\frac{2x-5}{4}=\frac{x+1}{3}\)
b, \(\frac{x-\frac{1}{2}}{2^3-3^3}=\frac{8x+1}{16^2-2^7}\)
nói cách làm nữa nhé
Tìm x biết
a, |x-1| <,= 2
b, |2x+5| >,= \(\frac{1}{2}\)
nói cách làm nữa nah
Mọi người giúp tới gấp nhé:
1. Tìm x, biết:
a/ 3(2x - 3) + 2(2 - x) = -3
b/ 2x(x2 - 2) + x2(1 - 2x) - x2 = -12
2. Tìm x, biết:
a/ 3x(2x + 3) - (2x + 5)(3x - 2) = 8
b/ 4x(x - 1) - 3(x2 - 5) - x2 = (x - 3) - (x + 4)
c/ 2(3x - 1)(2x + 5) - 6(2x - 1)(x + 2) = -6
d/ 3(2x - 1)(3x - 1) - (2x - 3)(9x -1) - 3 = -3
e/ (3x - 1)(2x + 7) - (x + 1)(6x - 5) = (x + 2) - (x - 5)
f/ 3xy(x + y) - (x + y)(x2 + y2 + 2xy) + y3 = 27
3. Chứng minh rằng giá trị của các biểu thức sau không phụ thuộc vào x:
a/ A = 2x(x - 1) - x(2x + 1) - (3 - 3x)
b/ B = 2x(x - 3) - (2x - 2)(x - 2)
c/ C = (3x - 5)(2x + 11) - (2x + 3)(3x + 7)
d/ D = (2x + 11)(3x - 5) - (2x + 3)(3x + 7)
f/ \(3xy\left(x+y\right)-\left(x+y\right)\left(x^2+y^2+2xy\right)+y^3=27\)
\(3x^2y+3xy^2-\left(x+y\right)\left(x+y\right)^2+y^3=27\)
\(3x^2y+3xy^3-\left(x+y\right)^3+y^3=27\)
\(3x^2y+3xy^3-\left(x^3+3x^2y+3xy^2+b^3\right)+y^3=27\)
\(-x^3=27\)
\(x=-3\)
Bài 1:
a/ \(3\left(2x-3\right)+2\left(2-x\right)=-3\)
\(6x-9+4-2x=-3\)
\(4x=-2\)
\(x=-\frac{1}{2}\)
b/ \(2x\left(x^2-2\right)+x^2\left(1-2x\right)-x^2=-12\)
\(2x^3-4x+x^2-2x^3-x^2=-12\)
\(-4x=-12\)
\(x=\frac{1}{3}\)
Bài 2:
a/ \(3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\)
\(6x^2+9x-6x^2-15x+4x+10=8\)
\(-2x=8\)
\(x=-4\)
b/ \(4x\left(x-1\right)-3\left(x^2-5\right)-x^2=\left(x-3\right)-\left(x+4\right)\)
\(4x^2-4x-3x^2+15-x^2=-7\)
\(-4x=-22\)
\(x=\frac{11}{2}\)
c/ \(2\left(3x-1\right)\left(2x+5\right)-6\left(2x-1\right)\left(x+2\right)=-6\)
\(6x-2\left(2x+5\right)-12x+6\left(x+2\right)=-6\)
\(6x-4x-10-12x+6x+12=-6\)
\(-4x=-8\)
\(x=2\)
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
1.Tìm x , biết
a.1/2x + 5/2=7/2x - 3/4
b.2/3x-2/5=1/2x-1/3
c.1/3x + 2/5 ( x+1) = 0
d.2/3-1/3(x-3/2) -1/2 (2x+1) = 5
e. 11/15 -( 7/9 +x ) * 3/8=61/90+x/3
j.5x(2x-1/2)+2(2x-1/2)= 0
Các bạn giải giúp mình nhé .Giải chi tiết nhé . Mik đang cần gấp . Cảm ơn
\(a,\frac{1}{2}x+\frac{5}{2}=\frac{7}{2}x-\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{2}x+\frac{5}{2}-\frac{7}{2}x=-\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{2}x-\frac{7}{2}x+\frac{5}{2}=-\frac{3}{4}\)
\(\Leftrightarrow-3x+\frac{5}{2}=-\frac{3}{4}\)
\(\Leftrightarrow-3x=-\frac{13}{4}\)
\(\Leftrightarrow x=-\frac{13}{4}:(-3)=-\frac{13}{4}:\frac{-3}{1}=-\frac{13}{4}\cdot\frac{-1}{3}=\frac{13}{12}\)
\(b,\frac{2}{3}x-\frac{2}{5}=\frac{1}{2}x-\frac{1}{3}\)
\(\Leftrightarrow\frac{2}{3}x-\frac{2}{5}-\frac{1}{2}x=-\frac{1}{3}\)
\(\Leftrightarrow\frac{2}{3}x-\frac{1}{2}x-\frac{2}{5}=-\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{6}x-\frac{2}{5}=-\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{6}x=\frac{1}{15}\)
\(\Leftrightarrow x=\frac{1}{15}:\frac{1}{6}=\frac{1}{15}\cdot6=\frac{6}{15}=\frac{2}{5}\)
\(c,\frac{1}{3}x+\frac{2}{5}(x+1)=0\)
\(\Leftrightarrow\frac{1}{3}x+\frac{2}{5}x+\frac{2}{5}=0\)
\(\Leftrightarrow\frac{11}{15}x=-\frac{2}{5}\)
\(\Leftrightarrow x=-\frac{6}{11}\)
d,e,f Tương tự